{"id":1656,"date":"2015-04-24T16:44:24","date_gmt":"2015-04-24T16:44:24","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=1656"},"modified":"2015-08-28T16:57:58","modified_gmt":"2015-08-28T16:57:58","slug":"molarity-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/molarity-2\/","title":{"raw":"Molarity","rendered":"Molarity"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>LEARNING OBJECTIVES<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n\t<li>Describe the fundamental properties of solutions<\/li>\r\n\t<li>Calculate solution concentrations using molarity<\/li>\r\n\t<li>Perform dilution calculations using the dilution equation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm232512\">In preceding sections, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures\u2014samples of matter containing two or more substances physically combined\u2014are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet\u2019s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an \u201calloy\u201d) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 1). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"501\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211200\/CNX_Chem_03_03_espresso1.jpg\" alt=\"A picture is shown of sugar being poured from a spoon into a cup.\" width=\"501\" height=\"439\" data-media-type=\"image\/jpeg\" \/> Figure 1. Sugar is one of many components in the complex mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the beverage\u2019s sweetness. (credit: Jane Whitney)[\/caption]\r\n\r\n<section id=\"fs-idm10227280\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Solutions<\/h2>\r\n<p id=\"fs-idm60561504\">We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions.<\/p>\r\n<p id=\"fs-idm635664\">The relative amount of a given solution component is known as its <strong><span data-type=\"term\">concentration<\/span><\/strong>. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the <strong><span data-type=\"term\">solvent<\/span><\/strong> and may be viewed as the medium in which the other components are dispersed, or <strong><span data-type=\"term\">dissolved<\/span><\/strong>. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an<strong> <span data-type=\"term\">aqueous solution<\/span><\/strong>.<\/p>\r\n<p id=\"fs-idm64279024\">A <strong><span data-type=\"term\">solute<\/span><\/strong> is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as <strong><span data-type=\"term\">dilute<\/span><\/strong> (of relatively low concentration) and <strong><span data-type=\"term\">concentrated<\/span><\/strong> (of relatively high concentration).<\/p>\r\n<p id=\"fs-idm62212352\">Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. <strong><span data-type=\"term\">Molarity (<em data-effect=\"italics\">M<\/em>)<\/span> <\/strong>is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:<\/p>\r\n\r\n<div id=\"fs-idm57520096\" data-type=\"equation\">[latex]M=\\frac{\\text{mol solute}}{\\text{L solution}}[\/latex]<\/div>\r\n<div id=\"fs-idm98982768\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1<\/h3>\r\n<h4 id=\"fs-idm10181424\"><strong><span data-type=\"title\">Calculating Molar Concentrations<\/span><\/strong><\/h4>\r\nA 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?\r\n<h4 id=\"fs-idm60197152\"><strong><span data-type=\"title\">Solution<\/span><\/strong><\/h4>\r\nSince the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:\r\n<div id=\"fs-idm77939616\" data-type=\"equation\">[latex]M=\\frac{\\text{mol solute}}{\\text{L solution}}=\\frac{0.133\\text{mol}}{355\\text{mL}\\times \\frac{1\\text{L}}{1000\\text{mL}}}=0.375M[\/latex]<\/div>\r\n<h4 id=\"fs-idm883648\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?\r\n<div id=\"fs-idm62768528\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a00.05 <em data-effect=\"italics\">M<\/em><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2<\/h3>\r\n<h4 id=\"fs-idm58170960\"><strong><span data-type=\"title\">Deriving Moles and Volumes from Molar Concentrations<\/span><\/strong><\/h4>\r\nHow much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example 1?\r\n<h4 id=\"fs-idm768\"><strong><span data-type=\"title\">Solution<\/span><\/strong><\/h4>\r\nIn this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 1, 0.375 <em data-effect=\"italics\">M<\/em>:\r\n<div id=\"fs-idm72962928\" data-type=\"equation\">[latex]\\begin{array}{c}\\\\ M=\\frac{\\text{mol solute}}{\\text{L solution}}\\\\ \\text{mol solute}=M\\times \\text{L solution}\\\\ \\\\ \\text{mol solute}=0.375\\frac{\\text{mol sugar}}{\\text{L}}\\times \\left(10\\text{mL}\\times \\frac{1\\text{L}}{1000\\text{mL}}\\right)=0.004\\text{mol sugar}\\end{array}[\/latex]<\/div>\r\n<h4 id=\"fs-idm97768960\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nWhat volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?\r\n<div id=\"fs-idm58199648\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a080 mL<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm64107376\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3<\/h3>\r\n<h4 id=\"fs-idm98918048\"><strong><span data-type=\"title\">Calculating Molar Concentrations from the Mass of Solute<\/span><\/strong><\/h4>\r\nDistilled white vinegar (Figure 2) is a solution of acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"500\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211201\/CNX_Chem_03_04_vinegar1.jpg\" alt=\"A label on a container is shown. The label has a picture of a salad with the words \u201cDistilled White Vinegar,\u201d and, \u201cReduced with water to 5% acidity,\u201d written above it.\" width=\"500\" height=\"476\" data-media-type=\"image\/jpeg\" \/> Figure 2. Distilled white vinegar is a solution of acetic acid in water.[\/caption]\r\n<h4 id=\"fs-idm82387696\"><strong><span data-type=\"title\">Solution<\/span><\/strong><\/h4>\r\nAs in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute\u2019s molar mass to obtain the amount of solute in moles:\r\n<div id=\"fs-idm81742128\" data-type=\"equation\">[latex]M=\\frac{\\text{mol solute}}{\\text{L solution}}=\\frac{25.2 g{\\text{CH}}_{2}{\\text{CO}}_{2}\\text{H}\\times \\frac{1{\\text{mol CH}}_{2}{\\text{CO}}_{2}\\text{H}}{{\\text{60.052 g CH}}_{2}{\\text{CO}}_{2}\\text{H}}}{\\text{0.500 L solution}}=0.839M[\/latex]<\/div>\r\n<div id=\"fs-idm8804720\" data-type=\"equation\">[latex]\\begin{array}{l}\\\\ M=\\frac{\\text{mol solute}}{\\text{L solution}}=0.839M\\\\ M=\\frac{0.839\\text{mol solute}}{1.00\\text{L solution}}\\end{array}[\/latex]<\/div>\r\n<h4 id=\"fs-idm61725600\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nCalculate the molarity of 6.52 g of CoCl<sub>2<\/sub> (128.9 g\/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.\r\n<div id=\"fs-idm2856432\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:0.674 <em data-effect=\"italics\">M<\/em><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm104693104\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4<\/h3>\r\n<h4 id=\"fs-idm2448752\"><strong><span data-type=\"title\">Determining the Mass of Solute in a Given Volume of Solution<\/span><\/strong><\/h4>\r\nHow many grams of NaCl are contained in 0.250 L of a 5.30-<em data-effect=\"italics\">M<\/em> solution?\r\n<h4 id=\"fs-idm122143856\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nThe volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 2:\r\n<div id=\"fs-idm85599968\" data-type=\"equation\">[latex]\\begin{array}{c}\\\\ M=\\frac{\\text{mol solute}}{\\text{L solution}}\\\\ \\text{mol solute}=M\\times \\text{L solution}\\\\ \\text{mol solute}=5.30\\frac{\\text{mol NaCl}}{\\text{L}}\\times 0.250\\text{L}=1.325\\text{mol NaCl}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idm108465360\">Finally, this molar amount is used to derive the mass of NaCl:<\/p>\r\n\r\n<div id=\"fs-idm1818176\" data-type=\"equation\">[latex]\\text{1.325 mol NaCl}\\times \\frac{58.44\\text{g NaCl}}{\\text{mol NaCl}}=77.4\\text{g NaCl}[\/latex]<\/div>\r\n<h4 id=\"fs-idm122705296\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nHow many grams of CaCl<sub>2<\/sub> (110.98 g\/mol) are contained in 250.0 mL of a 0.200-<em data-effect=\"italics\">M<\/em> solution of calcium chloride?\r\n<div id=\"fs-idm80099776\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a05.55 g CaCl<sub>2<\/sub><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm75726960\">When performing calculations stepwise, as in Example 4,\u00a0it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.<\/p>\r\n<p id=\"fs-idm105202592\">In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (see Example 5). This eliminates intermediate steps so that only the final result is rounded.<\/p>\r\n\r\n<div id=\"fs-idm88061984\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5<\/h3>\r\n<h4 id=\"fs-idm111967328\"><strong><span data-type=\"title\">Determining the Volume of Solution Containing a Given Mass of Solute<\/span><\/strong><\/h4>\r\nIn Example 3, we found the typical concentration of vinegar to be 0.839 <em data-effect=\"italics\">M<\/em>. What volume of vinegar contains 75.6 g of acetic acid?\r\n<h4 id=\"fs-idm108289568\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nFirst, use the molar mass to calculate moles of acetic acid from the given mass:\r\n<div id=\"fs-idm112801328\" data-type=\"equation\">[latex]\\text{g solute}\\times \\frac{\\text{mol solute}}{\\text{g solute}}=\\text{mol solute}[\/latex]<\/div>\r\n<p id=\"fs-idm67563520\">Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:<\/p>\r\n\r\n<div id=\"fs-idm79868224\" data-type=\"equation\">[latex]\\text{mol solute}\\times \\frac{\\text{L solution}}{\\text{mol solute}}=\\text{L solution}[\/latex]<\/div>\r\n<p id=\"fs-idm141610240\">Combining these two steps into one yields:<\/p>\r\n\r\n<div id=\"fs-idm104886656\" data-type=\"equation\">[latex]\\text{g solute}\\times \\frac{\\text{mol solute}}{\\text{g solute}}\\times \\frac{\\text{L solution}}{\\text{mol solute}}=\\text{L solution}[\/latex]<\/div>\r\n<div id=\"fs-idm79860368\" data-type=\"equation\">[latex]75.6\\text{g}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(\\frac{\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}}{60.05\\text{g}}\\right)\\left(\\frac{\\text{L solution}}{0.839\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}}\\right)=1.50\\text{L solution}[\/latex]<\/div>\r\n<h4 id=\"fs-idm59437344\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nWhat volume of a 1.50-<em data-effect=\"italics\">M<\/em> KBr solution contains 66.0 g KBr?\r\n<div id=\"fs-idm87238064\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a00.370 L<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-idm97949648\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Dilution of Solutions<\/h2>\r\n<p id=\"fs-idm134700400\"><span data-type=\"term\">Dilution<\/span> is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 3).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"500\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211203\/CNX_Chem_03_04_dilution1.jpg\" alt=\"This figure shows two graduated cylinders side-by-side. The first has about half as much blue liquid as the second. The blue liquid is darker in the first cylinder than in the second.\" width=\"500\" height=\"296\" data-media-type=\"image\/jpeg\" \/> Figure 3. Both solutions contain the same mass of copper nitrate. The solution on the right is more dilute because the copper nitrate is dissolved in more solvent. (credit: Mark Ott)[\/caption]\r\n<p id=\"fs-idm77995840\">Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated <em data-effect=\"italics\">stock solution<\/em>, we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure 4).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"881\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211204\/CNX_Chem_03_04_solution1.jpg\" alt=\"This figure shows two photos. In the first, there is an empty glass container, 4.75 g of K M n O subscript 4 powder on a white circle, and a bottle of distilled water. In the second photo the powder and about half the water have been added to the glass container. The liquid in the glass container is almost black in color.\" width=\"881\" height=\"366\" data-media-type=\"image\/jpeg\" \/> Figure 4. A solution of KMnO4 is prepared by mixing water with 4.74 g of KMnO4 in a flask. (credit: modification of work by Mark Ott)[\/caption]\r\n<p id=\"fs-idm77996832\">A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution\u2019s molarity and its volume in liters:<\/p>\r\n\r\n<div id=\"fs-idm32193760\" data-type=\"equation\">[latex]n=ML[\/latex]<\/div>\r\n<p id=\"fs-idm104760288\">Expressions like these may be written for a solution before and after it is diluted:<\/p>\r\n\r\n<div id=\"fs-idm18261728\" data-type=\"equation\">[latex]{n}_{1}={M}_{1}{L}_{1}[\/latex]<\/div>\r\n<div id=\"fs-idm61402048\" data-type=\"equation\">[latex]{n}_{2}={M}_{2}{L}_{2}[\/latex]<\/div>\r\n<p id=\"fs-idm123215808\">where the subscripts \u201c1\u201d and \u201c2\u201d refer to the solution before and after the dilution, respectively. Since the dilution process <em data-effect=\"italics\">does not change the amount of solute in the solution,<\/em><em data-effect=\"italics\">n<\/em><sub>1<\/sub> = <em data-effect=\"italics\">n<\/em><sub>2<\/sub>. Thus, these two equations may be set equal to one another:<\/p>\r\n\r\n<div id=\"fs-idp188032944\" data-type=\"equation\">[latex]{M}_{1}{L}_{1}={M}_{2}{L}_{2}[\/latex]<\/div>\r\n<p id=\"fs-idm103311152\">This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:<\/p>\r\n\r\n<div id=\"fs-idm69146864\" data-type=\"equation\">[latex]{C}_{1}{V}_{1}={C}_{2}{V}_{2}[\/latex]<\/div>\r\n<p id=\"fs-idm81143040\">where <em data-effect=\"italics\">C<\/em> and <em data-effect=\"italics\">V<\/em> are concentration and volume, respectively.<\/p>\r\n\r\n<div id=\"fs-idm98324208\" class=\"chemistry link-to-learning textbox\" data-type=\"note\">\r\n<p style=\"text-align: left;\">Use the <a href=\"https:\/\/phet.colorado.edu\/en\/simulation\/concentration\" target=\"_blank\">PhET simulation\u00a0for Concentration<\/a>\u00a0to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm81737600\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example\u00a06<\/h3>\r\n<h4 id=\"fs-idm65542512\"><strong><span data-type=\"title\">Determining the Concentration of a Diluted Solution<\/span><\/strong><\/h4>\r\nIf 0.850 L of a 5.00-<em data-effect=\"italics\">M<\/em> solution of copper nitrate, Cu(NO<sub>3<\/sub>)<sub>2<\/sub>, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?\r\n<h4 id=\"fs-idm63250768\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nWe are given the volume and concentration of a stock solution, <em data-effect=\"italics\">V<\/em><sub>1<\/sub> and <em data-effect=\"italics\">C<\/em><sub>1<\/sub>, and the volume of the resultant diluted solution, <em data-effect=\"italics\">V<\/em><sub>2<\/sub>. We need to find the concentration of the diluted solution, <em data-effect=\"italics\">C<\/em><sub>2<\/sub>. We thus rearrange the dilution equation in order to isolate <em data-effect=\"italics\">C<\/em><sub>2<\/sub>:\r\n<div id=\"fs-idm82973104\" data-type=\"equation\">[latex]\\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\\\ \\\\ {C}_{2}=\\frac{{C}_{1}{V}_{1}}{{V}_{2}}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idm81759792\">Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution\u2019s concentration to be less than one-half 5 <em data-effect=\"italics\">M<\/em>. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:<\/p>\r\n\r\n<div id=\"fs-idm61572480\" data-type=\"equation\">[latex]{C}_{2}=\\frac{0.850\\text{L}\\times 5.00\\frac{\\text{mol}}{\\text{L}}}{1.80 L}=2.36M[\/latex]<\/div>\r\n<p id=\"fs-idm125598048\">This result compares well to our ballpark estimate (it\u2019s a bit less than one-half the stock concentration, 5 <em data-effect=\"italics\">M<\/em>).<\/p>\r\n\r\n<h4 id=\"fs-idm105016944\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nWhat is the concentration of the solution that results from diluting 25.0 mL of a 2.04-<em data-effect=\"italics\">M<\/em> solution of CH<sub>3<\/sub>OH to 500.0 mL?\r\n<div id=\"fs-idm112503952\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a00.102 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>OH<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7<\/h3>\r\n<h4 id=\"fs-idm90576336\"><strong><span data-type=\"title\">Volume of a Diluted Solution<\/span><\/strong><\/h4>\r\nWhat volume of 0.12 <em data-effect=\"italics\">M<\/em> HBr can be prepared from 11 mL (0.011 L) of 0.45 <em data-effect=\"italics\">M<\/em> HBr?\r\n<h4 id=\"fs-idm75791728\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nWe are given the volume and concentration of a stock solution, <em data-effect=\"italics\">V<\/em><sub>1<\/sub> and <em data-effect=\"italics\">C<\/em><sub>1<\/sub>, and the concentration of the resultant diluted solution, <em data-effect=\"italics\">C<\/em><sub>2<\/sub>. We need to find the volume of the diluted solution, <em data-effect=\"italics\">V<\/em><sub>2<\/sub>. We thus rearrange the dilution equation in order to isolate <em data-effect=\"italics\">V<\/em><sub>2<\/sub>:\r\n<div id=\"fs-idm80850240\" data-type=\"equation\">[latex]\\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\\\ \\\\ {V}_{2}=\\frac{{C}_{1}{V}_{1}}{{C}_{2}}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idm61209504\">Since the diluted concentration (0.12 <em data-effect=\"italics\">M<\/em>) is slightly more than one-fourth the original concentration (0.45 <em data-effect=\"italics\">M<\/em>), we would expect the volume of the diluted solution to be roughly four times the original concentration, or around 44 mL. Substituting the given values and solving for the unknown volume yields:<\/p>\r\n\r\n<div id=\"fs-idm82787888\" data-type=\"equation\">[latex]\\begin{array}{c}\\\\ {V}_{2}=\\frac{\\left(0.45M\\right)\\left(0.011\\text{L}\\right)}{\\left(0.12M\\right)}\\\\ {V}_{2}=0.041\\text{L}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idm80652096\">The volume of the 0.12-<em data-effect=\"italics\">M<\/em> solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate.<\/p>\r\n\r\n<h4 id=\"fs-idm78684320\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA laboratory experiment calls for 0.125 <em data-effect=\"italics\">M<\/em> HNO<sub>3<\/sub>. What volume of 0.125 <em data-effect=\"italics\">M<\/em> HNO<sub>3<\/sub> can be prepared from 0.250 L of 1.88 <em data-effect=\"italics\">M<\/em> HNO<sub>3<\/sub>?\r\n<div id=\"fs-idm58675056\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>: \u00a03.76 L<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm81033056\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8<\/h3>\r\n<h4 id=\"fs-idm58713072\"><strong><span data-type=\"title\">Volume of a Concentrated Solution Needed for Dilution<\/span><\/strong><\/h4>\r\nWhat volume of 1.59 <em data-effect=\"italics\">M<\/em> KOH is required to prepare 5.00 L of 0.100 <em data-effect=\"italics\">M<\/em> KOH?\r\n<h4 id=\"fs-idm72581456\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nWe are given the concentration of a stock solution, <em data-effect=\"italics\">C<\/em><sub>1<\/sub>, and the volume and concentration of the resultant diluted solution, <em data-effect=\"italics\">V<\/em><sub>2<\/sub> and <em data-effect=\"italics\">C<\/em><sub>2<\/sub>. We need to find the volume of the stock solution, <em data-effect=\"italics\">V<\/em><sub>1<\/sub>. We thus rearrange the dilution equation in order to isolate <em data-effect=\"italics\">V<\/em><sub>1<\/sub>:\r\n<div id=\"fs-idm108330752\" data-type=\"equation\">[latex]\\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\\\ \\\\ {V}_{1}=\\frac{{C}_{2}{V}_{2}}{{C}_{1}}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idm108418736\">Since the concentration of the diluted solution 0.100 <em data-effect=\"italics\">M<\/em> is roughly one-sixteenth that of the stock solution (1.59 <em data-effect=\"italics\">M<\/em>), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:<\/p>\r\n\r\n<div id=\"fs-idm129889216\" data-type=\"equation\">[latex]\\begin{array}{c}\\\\ {V}_{1}=\\frac{\\left(0.100M\\right)\\left(5.00\\text{L}\\right)}{1.59M}\\\\ {V}_{1}=0.314\\text{L}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idm85673088\">Thus, we would need 0.314 L of the 1.59-<em data-effect=\"italics\">M<\/em> solution to prepare the desired solution. This result is consistent with our rough estimate.<\/p>\r\n\r\n<h4 id=\"fs-idm61586320\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nWhat volume of a 0.575-<em data-effect=\"italics\">M<\/em> solution of glucose, C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>, can be prepared from 50.00 mL of a 3.00-<em data-effect=\"italics\">M<\/em> glucose solution?\r\n<div id=\"fs-idm150115024\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>: \u00a00.261 L<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm60422464\" data-type=\"example\"><section id=\"fs-idm40478192\" class=\"summary\" data-depth=\"1\">\r\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\r\n<div class=\"entry-content\">\r\n<div class=\"im_section\">\r\n<div class=\"im_section\">\r\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<section>\r\n<div data-type=\"note\">\r\n<p id=\"fs-idp236281408\">Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Equations<\/h3>\r\n<section>\r\n<div data-type=\"note\">\r\n<ul>\r\n\t<li>[latex]M=\\frac{\\text{mol solute}}{\\text{L solution}}[\/latex]<\/li>\r\n\t<li><em data-effect=\"italics\">C<\/em><sub>1<\/sub><em data-effect=\"italics\">V<\/em><sub>1<\/sub> = <em data-effect=\"italics\">C<\/em><sub>2<\/sub><em data-effect=\"italics\">V<\/em><sub>2<\/sub><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<div id=\"fs-idm107608448\" data-type=\"exercise\">\r\n<div id=\"fs-idm106693616\" data-type=\"problem\">\r\n<ol>\r\n\t<li id=\"fs-idm100698576\">Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.<\/li>\r\n\t<li>What information do we need to calculate the molarity of a sulfuric acid solution?<\/li>\r\n\t<li>What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different<\/li>\r\n\t<li>Determine the molarity for each of the following solutions:\r\n<ol>\r\n\t<li>0.444 mol of CoCl<sub>2<\/sub> in 0.654 L of solution<\/li>\r\n\t<li>98.0 g of phosphoric acid, H<sub>3<\/sub>PO<sub>4<\/sub>, in 1.00 L of solution<\/li>\r\n\t<li>0.2074 g of calcium hydroxide, Ca(OH)<sub>2<\/sub>, in 40.00 mL of solution<\/li>\r\n\t<li>10.5 kg of Na<sub>2<\/sub>SO<sub>4<\/sub>\u202210H<sub>2<\/sub>O in 18.60 L of solution<\/li>\r\n\t<li>[latex]7.0\\times {10}^{-3}\\text{mol}[\/latex] of I<sub>2<\/sub> in 100.0 mL of solution<\/li>\r\n\t<li>[latex]1.8\\times {10}^{4}\\text{mg}[\/latex] of HCl in 0.075 L of solution<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Determine the molarity of each of the following solutions:\r\n<ol>\r\n\t<li>1.457 mol KCl in 1.500 L of solution<\/li>\r\n\t<li>0.515 g of H<sub>2<\/sub>SO<sub>4<\/sub> in 1.00 L of solution<\/li>\r\n\t<li>20.54 g of Al(NO<sub>3<\/sub>)<sub>3<\/sub> in 1575 mL of solution<\/li>\r\n\t<li>2.76 kg of CuSO<sub>4<\/sub>\u20225H<sub>2<\/sub>O in 1.45 L of solution<\/li>\r\n\t<li>0.005653 mol of Br<sub>2<\/sub> in 10.00 mL of solution<\/li>\r\n\t<li>0.000889 g of glycine, C<sub>2<\/sub>H<sub>5<\/sub>NO<sub>2<\/sub>, in 1.05 mL of solution<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Consider this question: What is the mass of the solute in 0.500 L of 0.30 <em data-effect=\"italics\">M<\/em> glucose, C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>, used for intravenous injection?\r\n<ol>\r\n\t<li>Outline the steps necessary to answer the question.<\/li>\r\n\t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Consider this question: What is the mass of solute in 200.0 L of a 1.556-<em data-effect=\"italics\">M<\/em> solution of KBr?\r\n<ol>\r\n\t<li>Outline the steps necessary to answer the question.<\/li>\r\n\t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Calculate the number of moles and the mass of the solute in each of the following solutions:\r\n<ol>\r\n\t<li>2.00 L of 18.5 <em data-effect=\"italics\">M<\/em> H<sub>2<\/sub>SO<sub>4<\/sub>, concentrated sulfuric acid<\/li>\r\n\t<li>100.0 mL of [latex]3.8\\times {10}^{-5}M\\text{NaCN,}[\/latex] the minimum lethal concentration of sodium cyanide in blood serum<\/li>\r\n\t<li>5.50 L of 13.3 <em data-effect=\"italics\">M<\/em> H<sub>2<\/sub>CO, the formaldehyde used to \u201cfix\u201d tissue samples<\/li>\r\n\t<li>325 mL of [latex]1.8\\times {10}^{-6}M[\/latex] FeSO<sub>4<\/sub>, the minimum concentration of iron sulfate detectable by taste in drinking water<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Calculate the number of moles and the mass of the solute in each of the following solutions:\r\n<ol>\r\n\t<li>325 mL of [latex]8.23\\times {10}^{-5}M\\text{KI}[\/latex], a source of iodine in the diet<\/li>\r\n\t<li>75.0 mL of [latex]2.2\\times {10}^{-5}M{\\text{H}}_{2}{\\text{SO}}_{4}[\/latex], a sample of acid rain<\/li>\r\n\t<li>0.2500 L of 0.1135 <em data-effect=\"italics\">M<\/em> K<sub>2<\/sub>CrO<sub>4<\/sub>, an analytical reagent used in iron assays<\/li>\r\n\t<li>10.5 L of 3.716 <em data-effect=\"italics\">M<\/em> (NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub>, a liquid fertilizer<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Consider this question: What is the molarity of KMnO<sub>4<\/sub> in a solution of 0.0908 g of KMnO<sub>4<\/sub> in 0.500 L of solution?\r\n<ol>\r\n\t<li>Outline the steps necessary to answer the question.<\/li>\r\n\t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl?\r\n<ol>\r\n\t<li>Outline the steps necessary to answer the question.<\/li>\r\n\t<li>Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Calculate the molarity of each of the following solutions:\r\n<ol>\r\n\t<li>0.195 g of cholesterol, C<sub>27<\/sub>H<sub>46<\/sub>O, in 0.100 L of serum, the average concentration of cholesterol in human serum<\/li>\r\n\t<li>4.25 g of NH<sub>3<\/sub> in 0.500 L of solution, the concentration of NH<sub>3<\/sub> in household ammonia<\/li>\r\n\t<li>1.49 kg of isopropyl alcohol, C<sub>3<\/sub>H<sub>7<\/sub>OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol<\/li>\r\n\t<li>0.029 g of I<sub>2<\/sub> in 0.100 L of solution, the solubility of I<sub>2<\/sub> in water at 20 \u00b0C<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Calculate the molarity of each of the following solutions:\r\n<ol>\r\n\t<li>293 g HCl in 666 mL of solution, a concentrated HCl solution<\/li>\r\n\t<li>2.026 g FeCl<sub>3<\/sub> in 0.1250 L of a solution used as an unknown in general chemistry laboratories<\/li>\r\n\t<li>0.001 mg Cd<sup>2+<\/sup> in 0.100 L, the maximum permissible concentration of cadmium in drinking water<\/li>\r\n\t<li>0.0079 g C<sub>7<\/sub>H<sub>5<\/sub>SNO<sub>3<\/sub> in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>There is about 1.0 g of calcium, as Ca<sup>2+<\/sup>, in 1.0 L of milk. What is the molarity of Ca<sup>2+<\/sup> in milk?<\/li>\r\n\t<li>What volume of a 1.00-<em data-effect=\"italics\">M<\/em> Fe(NO<sub>3<\/sub>)<sub>3<\/sub> solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 <em data-effect=\"italics\">M<\/em>?<\/li>\r\n\t<li>If 0.1718 L of a 0.3556-<em data-effect=\"italics\">M<\/em> C<sub>3<\/sub>H<sub>7<\/sub>OH solution is diluted to a concentration of 0.1222 <em data-effect=\"italics\">M<\/em>, what is the volume of the resulting solution?<\/li>\r\n\t<li>If 4.12 L of a 0.850 <em data-effect=\"italics\">M<\/em>-H<sub>3<\/sub>PO<sub>4<\/sub> solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution?<\/li>\r\n\t<li>What volume of a 0.33-<em data-effect=\"italics\">M<\/em> C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub> solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 <em data-effect=\"italics\">M<\/em>?<\/li>\r\n\t<li>What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-<em data-effect=\"italics\">M<\/em> solution is allowed to evaporate until the volume is reduced to 0.105 L?<\/li>\r\n\t<li>What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?\r\n<ol>\r\n\t<li>1.00 L of a 0.250-<em data-effect=\"italics\">M<\/em> solution of Fe(NO<sub>3<\/sub>)<sub>3<\/sub> is diluted to a final volume of 2.00 L<\/li>\r\n\t<li>0.5000 L of a 0.1222-<em data-effect=\"italics\">M<\/em> solution of C<sub>3<\/sub>H<sub>7<\/sub>OH is diluted to a final volume of 1.250 L<\/li>\r\n\t<li>2.35 L of a 0.350-<em data-effect=\"italics\">M<\/em> solution of H<sub>3<\/sub>PO<sub>4<\/sub> is diluted to a final volume of 4.00 L<\/li>\r\n\t<li>22.50 mL of a 0.025-<em data-effect=\"italics\">M<\/em> solution of C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub> is diluted to 100.0 mL<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>What is the final concentration of the solution produced when 225.5 mL of a 0.09988-<em data-effect=\"italics\">M<\/em> solution of Na<sub>2<\/sub>CO<sub>3<\/sub> is allowed to evaporate until the solution volume is reduced to 45.00 mL?<\/li>\r\n\t<li>A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?<\/li>\r\n\t<li>An experiment in a general chemistry laboratory calls for a 2.00-<em data-effect=\"italics\">M<\/em> solution of HCl. How many mL of 11.9 <em data-effect=\"italics\">M<\/em> HCl would be required to make 250 mL of 2.00 <em data-effect=\"italics\">M<\/em> HCl?<\/li>\r\n\t<li>What volume of a 0.20-<em data-effect=\"italics\">M<\/em> K<sub>2<\/sub>SO<sub>4<\/sub> solution contains 57 g of K<sub>2<\/sub>SO<sub>4<\/sub>?<\/li>\r\n\t<li>The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg\/L. If an industry is discharging hexavalent chromium as potassium dichromate (K<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub>), what is the maximum permissible molarity of that substance?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\r\n<div class=\"entry-content\">\r\n<div class=\"im_section\">\r\n<div class=\"im_section\">\r\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h4>Selected Answers<\/h4>\r\n2.\u00a0We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.\r\n\r\n4. (a) [latex]\\frac{0.444\\text{mol}}{0.654\\text{L}}=0.679{\\text{mol L}}^{-1}=0.679M\\text{;}[\/latex]\r\n<div id=\"fs-idm98103760\" data-type=\"exercise\">\r\n<div id=\"fs-idm10494144\" data-type=\"solution\">\r\n\r\n(b) First convert mass in grams to moles, and then substitute the proper terms into the definition.\r\n\r\nMolar mass of H<sub>3<\/sub>PO<sub>4<\/sub> = 97.995 g\/mol\r\n\r\n[latex]{\\text{mol (H}}_{3}{\\text{PO}}_{4}\\text{)}=98.0\\text{g}\\times \\frac{1\\text{mol}}{97.995\\text{g}}=1.00\\text{mol}[\/latex]\r\n<div data-type=\"newline\">[latex]M=\\frac{1.00\\text{mol}}{1.00\\text{L}}=1.00M\\text{;}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(c) Molar mass [Ca(OH)<sub>2<\/sub>] = 79.09 g\/mol<\/div>\r\n<div data-type=\"newline\">[latex]0.2074\\cancel{\\text{g}}\\times \\frac{\\text{1 mol}}{74.09\\cancel{\\text{g}}}=0.002799\\text{mol}{\\text{Ca(OH)}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\">[latex]\\frac{0.002799\\text{mol}}{0.0400\\text{L}}=0.06998{\\text{mol L}}^{-1}=0.06998M[\/latex] ;<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(d) Molar mass (Na<sub>2<\/sub>SO<sub>4<\/sub>\u202210H<sub>2<\/sub>O) = 322.20 g\/mol<\/div>\r\n<div data-type=\"newline\">[latex]10,500\\times \\frac{1\\text{mol}}{322.20\\text{g}}=32.6\\text{mol}[\/latex]<\/div>\r\n<div data-type=\"newline\">[latex]\\frac{32.6\\text{mol}}{18.60\\text{L}}=1.75M\\text{;}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(e) [latex]M=\\frac{\\text{millimoles solute}}{\\text{volume of solution in milliliters}}[\/latex]<\/div>\r\n<div data-type=\"newline\">[latex]\\frac{{\\text{7.00 mmol I}}_{2}}{\\text{100 mL}}=0.070M\\text{;}[\/latex] (f) Molar mass (HCl) = 36.46 g\/mol<\/div>\r\n<div data-type=\"newline\">[latex]\\text{mass (HCl)}=1.8\\times {10}^{1}\\text{g HCl}\\times \\frac{\\text{1 mol}}{\\text{36.46 g}}=\\text{0.49 mol HCl}[\/latex]<\/div>\r\n<div data-type=\"newline\">[latex]\\frac{\\text{0.49 mol HCl}}{\\text{0.075 L}}=\\text{6.6}M[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\"><\/div>\r\n<div id=\"fs-idm78757216\" data-type=\"exercise\">6. \u00a0(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass;<\/div>\r\n<div data-type=\"exercise\"><\/div>\r\n<div data-type=\"exercise\">(b) 0.500 L contains [latex]0.30M\\times 0.500\\text{L}=1.5\\times {10}^{-1}\\text{mol.}[\/latex] Molar mass (glucose): [latex]6\\times \\text{12.0011 g}+12\\times \\text{1.00794 g}+6\\times \\text{15.9994 g}=\\text{180.158 g},1.5\\times {10}^{-1}\\cancel{\\text{mol}}\\times \\text{180.158 g\/}\\cancel{\\text{mol}}=\\text{27 g.}[\/latex]<\/div>\r\n<div data-type=\"exercise\"><\/div>\r\n<div data-type=\"exercise\">8. The molarity must be converted to moles of solute, which is then converted to grams of solute:<span id=\"fs-idm106256160\" data-type=\"media\" data-alt=\"Three boxes connected by right-facing arrows in between each are shown. Written inside the boxes are the phrases, \u201cVolume of solution,\u201d \u201cmoles of solute,\u201d and \u201cmass of solute,\u201d respectively from left to right.\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211206\/CNX_Chem_03_04_Ex0408_img1.jpg\" alt=\"Three boxes connected by right-facing arrows in between each are shown. Written inside the boxes are the phrases, \u201cVolume of solution,\u201d \u201cmoles of solute,\u201d and \u201cmass of solute,\u201d respectively from left to right.\" width=\"884\" height=\"130\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<p id=\"fs-idm107786288\">[latex]M=\\frac{\\text{mol}}{\\text{liter}}\\text{or mol}=M\\times \\text{liter}[\/latex]<\/p>\r\n\r\n<div data-type=\"newline\">(a) [latex]\\begin{array}{l}\\\\ {\\text{mol H}}_{2}{\\text{SO}}_{4}=2.00\\cancel{\\text{L}}\\times \\frac{18.5\\text{mol}}{\\cancel{\\text{L}}}=37.0\\text{mol}{\\text{H}}_{2}{\\text{SO}}_{4}\\\\ 37.0\\cancel{{\\text{mol H}}_{2}{\\text{SO}}_{4}}\\times \\frac{98.08{\\text{g H}}_{2}{\\text{SO}}_{4}}{1\\cancel{{\\text{mol H}}_{2}{\\text{SO}}_{4}}}=3.63\\times {10}^{3}{\\text{g H}}_{2}{\\text{SO}}_{4}\\end{array}\\text{;}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(b) [latex]\\begin{array}{l}\\\\ \\text{mol NaCN}=0.1000\\cancel{\\text{L}}\\times \\frac{3.8\\times {10}^{-5}\\text{mol}}{\\cancel{\\text{L}}}=3.8\\times {10}^{-6}\\text{mol NaCN}\\\\ 3.8\\times {10}^{-5}\\cancel{\\text{mol NaCN}}\\times \\frac{49.01\\text{g}}{1\\cancel{\\text{mol NaCN}}}=1.9\\times {\\text{10}}^{-4}\\text{g NaCN}\\end{array}\\text{;}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(c)[latex]\\begin{array}{l}\\\\ {\\text{mol H}}_{2}\\text{CO}=5.50\\cancel{\\text{L}}\\times \\frac{13.3\\text{mol}}{\\cancel{\\text{L}}}=73.2\\text{mol}{\\text{H}}_{2}\\text{CO}\\\\ 73.2\\cancel{\\text{mol}{\\text{H}}_{2}\\text{CO}}\\times \\frac{30.026\\text{g}}{1\\cancel{{\\text{mol H}}_{2}\\text{CO}}}=2198\\text{g}{\\text{H}}_{2}\\text{CO}=2.20\\text{kg}{\\text{H}}_{2}\\text{CO}\\end{array}\\text{;}[\/latex]<\/div>\r\n(d) [latex]\\begin{array}{l}\\\\ {\\text{mol FeSO}}_{4}=0.325\\cancel{\\text{L}}\\times \\frac{1.8\\times {10}^{-6}\\text{mol}}{\\cancel{L}}=5.9\\times {10}^{-7}{\\text{mol FeSO}}_{4}\\\\ 5.85\\times {10}^{-7}\\cancel{{\\text{mol FeSO}}_{4}}\\times \\frac{151.9\\text{g}}{1\\cancel{{\\text{mol FeSO}}_{4}}}=8.9\\times {10}^{-5}{\\text{g FeSO}}_{4}\\end{array}[\/latex]\r\n\r\n&nbsp;\r\n\r\n10. (a) Determine the molar mass of KMnO<sub>4<\/sub>; determine the number of moles of KMnO<sub>4<\/sub> in the solution; from the number of moles and the volume of solution, determine the molarity;\r\n\r\n(b) Molar mass of KMnO<sub>4<\/sub> = 158.0264 g\/mol\r\n\r\n[latex]\\begin{array}{l}\\\\ \\text{mol}{\\text{KMnO}}_{4}=0.0908\\cancel{\\text{g}{\\text{KMnO}}_{4}}\\times \\frac{\\text{1 mol}}{158.0264\\cancel{{\\text{g KMnO}}_{4}}}=5.746\\times {10}^{-4}\\text{mol}\\\\ M{\\text{KMnO}}_{4}=\\frac{5.746\\times {10}^{-4}\\text{mol}}{0.500\\text{L}}=1.15\\times {10}^{-3}M\\end{array}[\/latex]\r\n\r\n&nbsp;\r\n\r\n12. (a) [latex]M{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}=\\frac{\\text{mol}}{V}=\\frac{\\frac{0.195\\cancel{g}{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}}{386.660\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}}}{0.100\\text{L}}=5.04\\times {10}^{-3}M\\text{;}[\/latex]\r\n\r\n(b) [latex]M{\\text{NH}}_{3}=\\frac{\\text{mol}}{V}=\\frac{\\frac{4.25\\cancel{g}{\\text{NH}}_{3}}{17.0304\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{NH}}_{3}}}{0.500\\text{L}}=0.499M\\text{;}[\/latex]\r\n\r\n(c) [latex]M{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}=\\frac{\\text{mol}}{V}=\\frac{1.49\\cancel{\\text{kg}}{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}\\times \\frac{1000\\cancel{g}}{1\\cancel{\\text{kg}}}\\times \\frac{1\\text{mol}{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}}{60.096\\cancel{\\text{g}}}}{2.50\\text{L}}=9.92M\\text{;}[\/latex]\r\n\r\n(d) [latex]M{\\text{I}}_{2}=\\frac{\\text{mol}}{V}=\\frac{\\frac{0.029\\cancel{\\text{g}}{\\text{I}}_{2}}{253.8090\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{I}}_{2}}}{0.100\\text{L}}=1.1\\times {10}^{-3}M[\/latex]\r\n\r\n&nbsp;\r\n\r\n14.\u00a0[latex]M=\\frac{\\text{mol}}{V}=\\frac{\\frac{1.0\\cancel{\\text{g}}}{40.08\\cancel{\\text{g}}{\\text{mol}}^{-1}}}{1.0\\text{L}}=0.025M[\/latex]\r\n\r\n16.\u00a0[latex]\\begin{array}{c}\\frac{{C}_{1}{V}_{1}}{{C}_{2}}={V}_{2}\\\\ \\frac{\\frac{0.3556\\text{mol}}{\\text{L}}\\times 0.1718\\text{L}}{\\frac{0.1222\\text{mol}}{\\text{L}}}={V}_{2}\\\\ 0.5000\\text{L}={V}_{2}\\end{array}[\/latex]\r\n\r\n18.\u00a0[latex]{V}_{1}=\\frac{{V}_{2}\\times {M}_{2}}{{M}_{2}}=25\\text{mL}\\times \\frac{0.025M}{0.33M}=1.9\\text{mL}[\/latex]\r\n\r\n20. (a) [latex]{C}_{2}=\\frac{{V}_{1}\\times {C}_{1}}{{V}_{2}}=1.00\\cancel{\\text{L}}\\times \\frac{0.250M}{2.00\\cancel{\\text{L}}}=0.125M\\text{;}[\/latex]\r\n<div data-type=\"exercise\">(b) [latex]{C}_{2}=\\frac{{V}_{1}\\times {C}_{1}}{{V}_{2}}=0.5000\\cancel{\\text{L}}\\times \\frac{0.1222M}{1.250\\cancel{\\text{L}}}=0.04888M\\text{;}[\/latex]<\/div>\r\n<div data-type=\"exercise\"><\/div>\r\n<div data-type=\"exercise\">(c) [latex]{C}_{2}=\\frac{{V}_{1}\\times {C}_{1}}{{V}_{2}}=2.35\\cancel{\\text{L}}\\times \\frac{0.350M}{4.00\\cancel{\\text{L}}}=0.206M\\text{;}[\/latex]<\/div>\r\n<div data-type=\"exercise\"><\/div>\r\n<div data-type=\"exercise\">(d) [latex]{C}_{2}=\\frac{{V}_{1}\\times {C}_{1}}{{V}_{2}}=0.02250\\cancel{\\text{mL}}\\times \\frac{0.025M}{0.100\\cancel{\\text{mL}}}=0.0056M[\/latex]<\/div>\r\n<div data-type=\"exercise\"><\/div>\r\n<div data-type=\"exercise\">22. Determine the number of moles in 434.4 g of HCl: 1.00794 + 35.4527 = 36.4606 g\/mol<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"fs-idm27373392\" data-type=\"exercise\">\r\n<div id=\"fs-idm27372432\" data-type=\"solution\">\r\n<div data-type=\"newline\">[latex]\\text{mol HCl}=\\frac{434.4\\cancel{g}}{36.4606\\cancel{g}{\\text{mol}}^{-1}}=11.91\\text{mol}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">This HCl is present in 1.00 L, so the molarity is 11.9 <em data-effect=\"italics\">M<\/em>.<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm27364240\" data-type=\"exercise\">\r\n\r\n&nbsp;\r\n\r\n24.\u00a0[latex]57\\text{g}{\\text{K}}_{2}{\\text{SO}}_{4}\\times \\frac{1\\text{mol}}{174.26\\text{g}}\\times \\frac{1\\text{L}}{0.20\\text{mol}}=1.6\\text{L}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"bcc-box bcc-success\"><section id=\"glossary\">\r\n<h3>Glossary<\/h3>\r\n<div data-type=\"definition\">\r\n<div id=\"fs-idm8143856\" data-type=\"definition\">\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">aqueous solution\r\n<\/span><\/strong>solution for which water is the solvent<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">concentrated\r\n<\/span><\/strong>qualitative term for a solution containing solute at a relatively high concentration<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">concentration\r\n<\/span><\/strong>quantitative measure of the relative amounts of solute and solvent present in a solution<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">dilute\r\n<\/span><\/strong>qualitative term for a solution containing solute at a relatively low concentration<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">dilution\r\n<\/span><\/strong>process of adding solvent to a solution in order to lower the concentration of solutes<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">dissolved\r\n<\/span><\/strong>describes the process by which solute components are dispersed in a solvent<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">molarity (<em data-effect=\"italics\">M<\/em>)\r\n<\/span><\/strong>unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution<\/p>\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">solute\r\n<\/span><\/strong>solution component present in a concentration less than that of the solvent<\/p>\r\n\r\n<div id=\"fs-idm26666032\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">solvent\r\n<\/span><\/strong>solution component present in a concentration that is higher relative to other components\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Describe the fundamental properties of solutions<\/li>\n<li>Calculate solution concentrations using molarity<\/li>\n<li>Perform dilution calculations using the dilution equation<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm232512\">In preceding sections, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures\u2014samples of matter containing two or more substances physically combined\u2014are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet\u2019s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an \u201calloy\u201d) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 1). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.<\/p>\n<div style=\"width: 511px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211200\/CNX_Chem_03_03_espresso1.jpg\" alt=\"A picture is shown of sugar being poured from a spoon into a cup.\" width=\"501\" height=\"439\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Sugar is one of many components in the complex mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the beverage\u2019s sweetness. (credit: Jane Whitney)<\/p>\n<\/div>\n<section id=\"fs-idm10227280\" data-depth=\"1\">\n<h2 data-type=\"title\">Solutions<\/h2>\n<p id=\"fs-idm60561504\">We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions.<\/p>\n<p id=\"fs-idm635664\">The relative amount of a given solution component is known as its <strong><span data-type=\"term\">concentration<\/span><\/strong>. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the <strong><span data-type=\"term\">solvent<\/span><\/strong> and may be viewed as the medium in which the other components are dispersed, or <strong><span data-type=\"term\">dissolved<\/span><\/strong>. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an<strong> <span data-type=\"term\">aqueous solution<\/span><\/strong>.<\/p>\n<p id=\"fs-idm64279024\">A <strong><span data-type=\"term\">solute<\/span><\/strong> is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as <strong><span data-type=\"term\">dilute<\/span><\/strong> (of relatively low concentration) and <strong><span data-type=\"term\">concentrated<\/span><\/strong> (of relatively high concentration).<\/p>\n<p id=\"fs-idm62212352\">Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. <strong><span data-type=\"term\">Molarity (<em data-effect=\"italics\">M<\/em>)<\/span> <\/strong>is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:<\/p>\n<div id=\"fs-idm57520096\" data-type=\"equation\">[latex]M=\\frac{\\text{mol solute}}{\\text{L solution}}[\/latex]<\/div>\n<div id=\"fs-idm98982768\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 1<\/h3>\n<h4 id=\"fs-idm10181424\"><strong><span data-type=\"title\">Calculating Molar Concentrations<\/span><\/strong><\/h4>\n<p>A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?<\/p>\n<h4 id=\"fs-idm60197152\"><strong><span data-type=\"title\">Solution<\/span><\/strong><\/h4>\n<p>Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:<\/p>\n<div id=\"fs-idm77939616\" data-type=\"equation\">[latex]M=\\frac{\\text{mol solute}}{\\text{L solution}}=\\frac{0.133\\text{mol}}{355\\text{mL}\\times \\frac{1\\text{L}}{1000\\text{mL}}}=0.375M[\/latex]<\/div>\n<h4 id=\"fs-idm883648\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?<\/p>\n<div id=\"fs-idm62768528\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a00.05 <em data-effect=\"italics\">M<\/em><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2<\/h3>\n<h4 id=\"fs-idm58170960\"><strong><span data-type=\"title\">Deriving Moles and Volumes from Molar Concentrations<\/span><\/strong><\/h4>\n<p>How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example 1?<\/p>\n<h4 id=\"fs-idm768\"><strong><span data-type=\"title\">Solution<\/span><\/strong><\/h4>\n<p>In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 1, 0.375 <em data-effect=\"italics\">M<\/em>:<\/p>\n<div id=\"fs-idm72962928\" data-type=\"equation\">[latex]\\begin{array}{c}\\\\ M=\\frac{\\text{mol solute}}{\\text{L solution}}\\\\ \\text{mol solute}=M\\times \\text{L solution}\\\\ \\\\ \\text{mol solute}=0.375\\frac{\\text{mol sugar}}{\\text{L}}\\times \\left(10\\text{mL}\\times \\frac{1\\text{L}}{1000\\text{mL}}\\right)=0.004\\text{mol sugar}\\end{array}[\/latex]<\/div>\n<h4 id=\"fs-idm97768960\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?<\/p>\n<div id=\"fs-idm58199648\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a080 mL<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm64107376\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 3<\/h3>\n<h4 id=\"fs-idm98918048\"><strong><span data-type=\"title\">Calculating Molar Concentrations from the Mass of Solute<\/span><\/strong><\/h4>\n<p>Distilled white vinegar (Figure 2) is a solution of acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?<\/p>\n<div style=\"width: 510px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211201\/CNX_Chem_03_04_vinegar1.jpg\" alt=\"A label on a container is shown. The label has a picture of a salad with the words \u201cDistilled White Vinegar,\u201d and, \u201cReduced with water to 5% acidity,\u201d written above it.\" width=\"500\" height=\"476\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Distilled white vinegar is a solution of acetic acid in water.<\/p>\n<\/div>\n<h4 id=\"fs-idm82387696\"><strong><span data-type=\"title\">Solution<\/span><\/strong><\/h4>\n<p>As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute\u2019s molar mass to obtain the amount of solute in moles:<\/p>\n<div id=\"fs-idm81742128\" data-type=\"equation\">[latex]M=\\frac{\\text{mol solute}}{\\text{L solution}}=\\frac{25.2 g{\\text{CH}}_{2}{\\text{CO}}_{2}\\text{H}\\times \\frac{1{\\text{mol CH}}_{2}{\\text{CO}}_{2}\\text{H}}{{\\text{60.052 g CH}}_{2}{\\text{CO}}_{2}\\text{H}}}{\\text{0.500 L solution}}=0.839M[\/latex]<\/div>\n<div id=\"fs-idm8804720\" data-type=\"equation\">[latex]\\begin{array}{l}\\\\ M=\\frac{\\text{mol solute}}{\\text{L solution}}=0.839M\\\\ M=\\frac{0.839\\text{mol solute}}{1.00\\text{L solution}}\\end{array}[\/latex]<\/div>\n<h4 id=\"fs-idm61725600\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>Calculate the molarity of 6.52 g of CoCl<sub>2<\/sub> (128.9 g\/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.<\/p>\n<div id=\"fs-idm2856432\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:0.674 <em data-effect=\"italics\">M<\/em><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm104693104\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 4<\/h3>\n<h4 id=\"fs-idm2448752\"><strong><span data-type=\"title\">Determining the Mass of Solute in a Given Volume of Solution<\/span><\/strong><\/h4>\n<p>How many grams of NaCl are contained in 0.250 L of a 5.30-<em data-effect=\"italics\">M<\/em> solution?<\/p>\n<h4 id=\"fs-idm122143856\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 2:<\/p>\n<div id=\"fs-idm85599968\" data-type=\"equation\">[latex]\\begin{array}{c}\\\\ M=\\frac{\\text{mol solute}}{\\text{L solution}}\\\\ \\text{mol solute}=M\\times \\text{L solution}\\\\ \\text{mol solute}=5.30\\frac{\\text{mol NaCl}}{\\text{L}}\\times 0.250\\text{L}=1.325\\text{mol NaCl}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idm108465360\">Finally, this molar amount is used to derive the mass of NaCl:<\/p>\n<div id=\"fs-idm1818176\" data-type=\"equation\">[latex]\\text{1.325 mol NaCl}\\times \\frac{58.44\\text{g NaCl}}{\\text{mol NaCl}}=77.4\\text{g NaCl}[\/latex]<\/div>\n<h4 id=\"fs-idm122705296\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>How many grams of CaCl<sub>2<\/sub> (110.98 g\/mol) are contained in 250.0 mL of a 0.200-<em data-effect=\"italics\">M<\/em> solution of calcium chloride?<\/p>\n<div id=\"fs-idm80099776\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a05.55 g CaCl<sub>2<\/sub><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idm75726960\">When performing calculations stepwise, as in Example 4,\u00a0it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.<\/p>\n<p id=\"fs-idm105202592\">In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (see Example 5). This eliminates intermediate steps so that only the final result is rounded.<\/p>\n<div id=\"fs-idm88061984\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 5<\/h3>\n<h4 id=\"fs-idm111967328\"><strong><span data-type=\"title\">Determining the Volume of Solution Containing a Given Mass of Solute<\/span><\/strong><\/h4>\n<p>In Example 3, we found the typical concentration of vinegar to be 0.839 <em data-effect=\"italics\">M<\/em>. What volume of vinegar contains 75.6 g of acetic acid?<\/p>\n<h4 id=\"fs-idm108289568\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>First, use the molar mass to calculate moles of acetic acid from the given mass:<\/p>\n<div id=\"fs-idm112801328\" data-type=\"equation\">[latex]\\text{g solute}\\times \\frac{\\text{mol solute}}{\\text{g solute}}=\\text{mol solute}[\/latex]<\/div>\n<p id=\"fs-idm67563520\">Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:<\/p>\n<div id=\"fs-idm79868224\" data-type=\"equation\">[latex]\\text{mol solute}\\times \\frac{\\text{L solution}}{\\text{mol solute}}=\\text{L solution}[\/latex]<\/div>\n<p id=\"fs-idm141610240\">Combining these two steps into one yields:<\/p>\n<div id=\"fs-idm104886656\" data-type=\"equation\">[latex]\\text{g solute}\\times \\frac{\\text{mol solute}}{\\text{g solute}}\\times \\frac{\\text{L solution}}{\\text{mol solute}}=\\text{L solution}[\/latex]<\/div>\n<div id=\"fs-idm79860368\" data-type=\"equation\">[latex]75.6\\text{g}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(\\frac{\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}}{60.05\\text{g}}\\right)\\left(\\frac{\\text{L solution}}{0.839\\text{mol}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}}\\right)=1.50\\text{L solution}[\/latex]<\/div>\n<h4 id=\"fs-idm59437344\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What volume of a 1.50-<em data-effect=\"italics\">M<\/em> KBr solution contains 66.0 g KBr?<\/p>\n<div id=\"fs-idm87238064\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a00.370 L<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idm97949648\" data-depth=\"1\">\n<h2 data-type=\"title\">Dilution of Solutions<\/h2>\n<p id=\"fs-idm134700400\"><span data-type=\"term\">Dilution<\/span> is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 3).<\/p>\n<div style=\"width: 510px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211203\/CNX_Chem_03_04_dilution1.jpg\" alt=\"This figure shows two graduated cylinders side-by-side. The first has about half as much blue liquid as the second. The blue liquid is darker in the first cylinder than in the second.\" width=\"500\" height=\"296\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Both solutions contain the same mass of copper nitrate. The solution on the right is more dilute because the copper nitrate is dissolved in more solvent. (credit: Mark Ott)<\/p>\n<\/div>\n<p id=\"fs-idm77995840\">Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated <em data-effect=\"italics\">stock solution<\/em>, we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure 4).<\/p>\n<div style=\"width: 891px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211204\/CNX_Chem_03_04_solution1.jpg\" alt=\"This figure shows two photos. In the first, there is an empty glass container, 4.75 g of K M n O subscript 4 powder on a white circle, and a bottle of distilled water. In the second photo the powder and about half the water have been added to the glass container. The liquid in the glass container is almost black in color.\" width=\"881\" height=\"366\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. A solution of KMnO4 is prepared by mixing water with 4.74 g of KMnO4 in a flask. (credit: modification of work by Mark Ott)<\/p>\n<\/div>\n<p id=\"fs-idm77996832\">A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution\u2019s molarity and its volume in liters:<\/p>\n<div id=\"fs-idm32193760\" data-type=\"equation\">[latex]n=ML[\/latex]<\/div>\n<p id=\"fs-idm104760288\">Expressions like these may be written for a solution before and after it is diluted:<\/p>\n<div id=\"fs-idm18261728\" data-type=\"equation\">[latex]{n}_{1}={M}_{1}{L}_{1}[\/latex]<\/div>\n<div id=\"fs-idm61402048\" data-type=\"equation\">[latex]{n}_{2}={M}_{2}{L}_{2}[\/latex]<\/div>\n<p id=\"fs-idm123215808\">where the subscripts \u201c1\u201d and \u201c2\u201d refer to the solution before and after the dilution, respectively. Since the dilution process <em data-effect=\"italics\">does not change the amount of solute in the solution,<\/em><em data-effect=\"italics\">n<\/em><sub>1<\/sub> = <em data-effect=\"italics\">n<\/em><sub>2<\/sub>. Thus, these two equations may be set equal to one another:<\/p>\n<div id=\"fs-idp188032944\" data-type=\"equation\">[latex]{M}_{1}{L}_{1}={M}_{2}{L}_{2}[\/latex]<\/div>\n<p id=\"fs-idm103311152\">This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:<\/p>\n<div id=\"fs-idm69146864\" data-type=\"equation\">[latex]{C}_{1}{V}_{1}={C}_{2}{V}_{2}[\/latex]<\/div>\n<p id=\"fs-idm81143040\">where <em data-effect=\"italics\">C<\/em> and <em data-effect=\"italics\">V<\/em> are concentration and volume, respectively.<\/p>\n<div id=\"fs-idm98324208\" class=\"chemistry link-to-learning textbox\" data-type=\"note\">\n<p style=\"text-align: left;\">Use the <a href=\"https:\/\/phet.colorado.edu\/en\/simulation\/concentration\" target=\"_blank\">PhET simulation\u00a0for Concentration<\/a>\u00a0to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.<\/p>\n<\/div>\n<div id=\"fs-idm81737600\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example\u00a06<\/h3>\n<h4 id=\"fs-idm65542512\"><strong><span data-type=\"title\">Determining the Concentration of a Diluted Solution<\/span><\/strong><\/h4>\n<p>If 0.850 L of a 5.00-<em data-effect=\"italics\">M<\/em> solution of copper nitrate, Cu(NO<sub>3<\/sub>)<sub>2<\/sub>, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?<\/p>\n<h4 id=\"fs-idm63250768\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>We are given the volume and concentration of a stock solution, <em data-effect=\"italics\">V<\/em><sub>1<\/sub> and <em data-effect=\"italics\">C<\/em><sub>1<\/sub>, and the volume of the resultant diluted solution, <em data-effect=\"italics\">V<\/em><sub>2<\/sub>. We need to find the concentration of the diluted solution, <em data-effect=\"italics\">C<\/em><sub>2<\/sub>. We thus rearrange the dilution equation in order to isolate <em data-effect=\"italics\">C<\/em><sub>2<\/sub>:<\/p>\n<div id=\"fs-idm82973104\" data-type=\"equation\">[latex]\\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\\\ \\\\ {C}_{2}=\\frac{{C}_{1}{V}_{1}}{{V}_{2}}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idm81759792\">Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution\u2019s concentration to be less than one-half 5 <em data-effect=\"italics\">M<\/em>. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:<\/p>\n<div id=\"fs-idm61572480\" data-type=\"equation\">[latex]{C}_{2}=\\frac{0.850\\text{L}\\times 5.00\\frac{\\text{mol}}{\\text{L}}}{1.80 L}=2.36M[\/latex]<\/div>\n<p id=\"fs-idm125598048\">This result compares well to our ballpark estimate (it\u2019s a bit less than one-half the stock concentration, 5 <em data-effect=\"italics\">M<\/em>).<\/p>\n<h4 id=\"fs-idm105016944\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What is the concentration of the solution that results from diluting 25.0 mL of a 2.04-<em data-effect=\"italics\">M<\/em> solution of CH<sub>3<\/sub>OH to 500.0 mL?<\/p>\n<div id=\"fs-idm112503952\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a00.102 <em data-effect=\"italics\">M<\/em> CH<sub>3<\/sub>OH<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7<\/h3>\n<h4 id=\"fs-idm90576336\"><strong><span data-type=\"title\">Volume of a Diluted Solution<\/span><\/strong><\/h4>\n<p>What volume of 0.12 <em data-effect=\"italics\">M<\/em> HBr can be prepared from 11 mL (0.011 L) of 0.45 <em data-effect=\"italics\">M<\/em> HBr?<\/p>\n<h4 id=\"fs-idm75791728\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>We are given the volume and concentration of a stock solution, <em data-effect=\"italics\">V<\/em><sub>1<\/sub> and <em data-effect=\"italics\">C<\/em><sub>1<\/sub>, and the concentration of the resultant diluted solution, <em data-effect=\"italics\">C<\/em><sub>2<\/sub>. We need to find the volume of the diluted solution, <em data-effect=\"italics\">V<\/em><sub>2<\/sub>. We thus rearrange the dilution equation in order to isolate <em data-effect=\"italics\">V<\/em><sub>2<\/sub>:<\/p>\n<div id=\"fs-idm80850240\" data-type=\"equation\">[latex]\\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\\\ \\\\ {V}_{2}=\\frac{{C}_{1}{V}_{1}}{{C}_{2}}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idm61209504\">Since the diluted concentration (0.12 <em data-effect=\"italics\">M<\/em>) is slightly more than one-fourth the original concentration (0.45 <em data-effect=\"italics\">M<\/em>), we would expect the volume of the diluted solution to be roughly four times the original concentration, or around 44 mL. Substituting the given values and solving for the unknown volume yields:<\/p>\n<div id=\"fs-idm82787888\" data-type=\"equation\">[latex]\\begin{array}{c}\\\\ {V}_{2}=\\frac{\\left(0.45M\\right)\\left(0.011\\text{L}\\right)}{\\left(0.12M\\right)}\\\\ {V}_{2}=0.041\\text{L}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idm80652096\">The volume of the 0.12-<em data-effect=\"italics\">M<\/em> solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate.<\/p>\n<h4 id=\"fs-idm78684320\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A laboratory experiment calls for 0.125 <em data-effect=\"italics\">M<\/em> HNO<sub>3<\/sub>. What volume of 0.125 <em data-effect=\"italics\">M<\/em> HNO<sub>3<\/sub> can be prepared from 0.250 L of 1.88 <em data-effect=\"italics\">M<\/em> HNO<sub>3<\/sub>?<\/p>\n<div id=\"fs-idm58675056\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>: \u00a03.76 L<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm81033056\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 8<\/h3>\n<h4 id=\"fs-idm58713072\"><strong><span data-type=\"title\">Volume of a Concentrated Solution Needed for Dilution<\/span><\/strong><\/h4>\n<p>What volume of 1.59 <em data-effect=\"italics\">M<\/em> KOH is required to prepare 5.00 L of 0.100 <em data-effect=\"italics\">M<\/em> KOH?<\/p>\n<h4 id=\"fs-idm72581456\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>We are given the concentration of a stock solution, <em data-effect=\"italics\">C<\/em><sub>1<\/sub>, and the volume and concentration of the resultant diluted solution, <em data-effect=\"italics\">V<\/em><sub>2<\/sub> and <em data-effect=\"italics\">C<\/em><sub>2<\/sub>. We need to find the volume of the stock solution, <em data-effect=\"italics\">V<\/em><sub>1<\/sub>. We thus rearrange the dilution equation in order to isolate <em data-effect=\"italics\">V<\/em><sub>1<\/sub>:<\/p>\n<div id=\"fs-idm108330752\" data-type=\"equation\">[latex]\\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\\\ \\\\ {V}_{1}=\\frac{{C}_{2}{V}_{2}}{{C}_{1}}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idm108418736\">Since the concentration of the diluted solution 0.100 <em data-effect=\"italics\">M<\/em> is roughly one-sixteenth that of the stock solution (1.59 <em data-effect=\"italics\">M<\/em>), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:<\/p>\n<div id=\"fs-idm129889216\" data-type=\"equation\">[latex]\\begin{array}{c}\\\\ {V}_{1}=\\frac{\\left(0.100M\\right)\\left(5.00\\text{L}\\right)}{1.59M}\\\\ {V}_{1}=0.314\\text{L}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idm85673088\">Thus, we would need 0.314 L of the 1.59-<em data-effect=\"italics\">M<\/em> solution to prepare the desired solution. This result is consistent with our rough estimate.<\/p>\n<h4 id=\"fs-idm61586320\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What volume of a 0.575-<em data-effect=\"italics\">M<\/em> solution of glucose, C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>, can be prepared from 50.00 mL of a 3.00-<em data-effect=\"italics\">M<\/em> glucose solution?<\/p>\n<div id=\"fs-idm150115024\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>: \u00a00.261 L<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm60422464\" data-type=\"example\">\n<section id=\"fs-idm40478192\" class=\"summary\" data-depth=\"1\">\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\n<div class=\"entry-content\">\n<div class=\"im_section\">\n<div class=\"im_section\">\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<section>\n<div data-type=\"note\">\n<p id=\"fs-idp236281408\">Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Equations<\/h3>\n<section>\n<div data-type=\"note\">\n<ul>\n<li>[latex]M=\\frac{\\text{mol solute}}{\\text{L solution}}[\/latex]<\/li>\n<li><em data-effect=\"italics\">C<\/em><sub>1<\/sub><em data-effect=\"italics\">V<\/em><sub>1<\/sub> = <em data-effect=\"italics\">C<\/em><sub>2<\/sub><em data-effect=\"italics\">V<\/em><sub>2<\/sub><\/li>\n<\/ul>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<div id=\"fs-idm107608448\" data-type=\"exercise\">\n<div id=\"fs-idm106693616\" data-type=\"problem\">\n<ol>\n<li id=\"fs-idm100698576\">Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.<\/li>\n<li>What information do we need to calculate the molarity of a sulfuric acid solution?<\/li>\n<li>What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different<\/li>\n<li>Determine the molarity for each of the following solutions:\n<ol>\n<li>0.444 mol of CoCl<sub>2<\/sub> in 0.654 L of solution<\/li>\n<li>98.0 g of phosphoric acid, H<sub>3<\/sub>PO<sub>4<\/sub>, in 1.00 L of solution<\/li>\n<li>0.2074 g of calcium hydroxide, Ca(OH)<sub>2<\/sub>, in 40.00 mL of solution<\/li>\n<li>10.5 kg of Na<sub>2<\/sub>SO<sub>4<\/sub>\u202210H<sub>2<\/sub>O in 18.60 L of solution<\/li>\n<li>[latex]7.0\\times {10}^{-3}\\text{mol}[\/latex] of I<sub>2<\/sub> in 100.0 mL of solution<\/li>\n<li>[latex]1.8\\times {10}^{4}\\text{mg}[\/latex] of HCl in 0.075 L of solution<\/li>\n<\/ol>\n<\/li>\n<li>Determine the molarity of each of the following solutions:\n<ol>\n<li>1.457 mol KCl in 1.500 L of solution<\/li>\n<li>0.515 g of H<sub>2<\/sub>SO<sub>4<\/sub> in 1.00 L of solution<\/li>\n<li>20.54 g of Al(NO<sub>3<\/sub>)<sub>3<\/sub> in 1575 mL of solution<\/li>\n<li>2.76 kg of CuSO<sub>4<\/sub>\u20225H<sub>2<\/sub>O in 1.45 L of solution<\/li>\n<li>0.005653 mol of Br<sub>2<\/sub> in 10.00 mL of solution<\/li>\n<li>0.000889 g of glycine, C<sub>2<\/sub>H<sub>5<\/sub>NO<sub>2<\/sub>, in 1.05 mL of solution<\/li>\n<\/ol>\n<\/li>\n<li>Consider this question: What is the mass of the solute in 0.500 L of 0.30 <em data-effect=\"italics\">M<\/em> glucose, C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>, used for intravenous injection?\n<ol>\n<li>Outline the steps necessary to answer the question.<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>Consider this question: What is the mass of solute in 200.0 L of a 1.556-<em data-effect=\"italics\">M<\/em> solution of KBr?\n<ol>\n<li>Outline the steps necessary to answer the question.<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the number of moles and the mass of the solute in each of the following solutions:\n<ol>\n<li>2.00 L of 18.5 <em data-effect=\"italics\">M<\/em> H<sub>2<\/sub>SO<sub>4<\/sub>, concentrated sulfuric acid<\/li>\n<li>100.0 mL of [latex]3.8\\times {10}^{-5}M\\text{NaCN,}[\/latex] the minimum lethal concentration of sodium cyanide in blood serum<\/li>\n<li>5.50 L of 13.3 <em data-effect=\"italics\">M<\/em> H<sub>2<\/sub>CO, the formaldehyde used to \u201cfix\u201d tissue samples<\/li>\n<li>325 mL of [latex]1.8\\times {10}^{-6}M[\/latex] FeSO<sub>4<\/sub>, the minimum concentration of iron sulfate detectable by taste in drinking water<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the number of moles and the mass of the solute in each of the following solutions:\n<ol>\n<li>325 mL of [latex]8.23\\times {10}^{-5}M\\text{KI}[\/latex], a source of iodine in the diet<\/li>\n<li>75.0 mL of [latex]2.2\\times {10}^{-5}M{\\text{H}}_{2}{\\text{SO}}_{4}[\/latex], a sample of acid rain<\/li>\n<li>0.2500 L of 0.1135 <em data-effect=\"italics\">M<\/em> K<sub>2<\/sub>CrO<sub>4<\/sub>, an analytical reagent used in iron assays<\/li>\n<li>10.5 L of 3.716 <em data-effect=\"italics\">M<\/em> (NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub>, a liquid fertilizer<\/li>\n<\/ol>\n<\/li>\n<li>Consider this question: What is the molarity of KMnO<sub>4<\/sub> in a solution of 0.0908 g of KMnO<sub>4<\/sub> in 0.500 L of solution?\n<ol>\n<li>Outline the steps necessary to answer the question.<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl?\n<ol>\n<li>Outline the steps necessary to answer the question.<\/li>\n<li>Answer the question.<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the molarity of each of the following solutions:\n<ol>\n<li>0.195 g of cholesterol, C<sub>27<\/sub>H<sub>46<\/sub>O, in 0.100 L of serum, the average concentration of cholesterol in human serum<\/li>\n<li>4.25 g of NH<sub>3<\/sub> in 0.500 L of solution, the concentration of NH<sub>3<\/sub> in household ammonia<\/li>\n<li>1.49 kg of isopropyl alcohol, C<sub>3<\/sub>H<sub>7<\/sub>OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol<\/li>\n<li>0.029 g of I<sub>2<\/sub> in 0.100 L of solution, the solubility of I<sub>2<\/sub> in water at 20 \u00b0C<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the molarity of each of the following solutions:\n<ol>\n<li>293 g HCl in 666 mL of solution, a concentrated HCl solution<\/li>\n<li>2.026 g FeCl<sub>3<\/sub> in 0.1250 L of a solution used as an unknown in general chemistry laboratories<\/li>\n<li>0.001 mg Cd<sup>2+<\/sup> in 0.100 L, the maximum permissible concentration of cadmium in drinking water<\/li>\n<li>0.0079 g C<sub>7<\/sub>H<sub>5<\/sub>SNO<sub>3<\/sub> in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink.<\/li>\n<\/ol>\n<\/li>\n<li>There is about 1.0 g of calcium, as Ca<sup>2+<\/sup>, in 1.0 L of milk. What is the molarity of Ca<sup>2+<\/sup> in milk?<\/li>\n<li>What volume of a 1.00-<em data-effect=\"italics\">M<\/em> Fe(NO<sub>3<\/sub>)<sub>3<\/sub> solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 <em data-effect=\"italics\">M<\/em>?<\/li>\n<li>If 0.1718 L of a 0.3556-<em data-effect=\"italics\">M<\/em> C<sub>3<\/sub>H<sub>7<\/sub>OH solution is diluted to a concentration of 0.1222 <em data-effect=\"italics\">M<\/em>, what is the volume of the resulting solution?<\/li>\n<li>If 4.12 L of a 0.850 <em data-effect=\"italics\">M<\/em>-H<sub>3<\/sub>PO<sub>4<\/sub> solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution?<\/li>\n<li>What volume of a 0.33-<em data-effect=\"italics\">M<\/em> C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub> solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 <em data-effect=\"italics\">M<\/em>?<\/li>\n<li>What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-<em data-effect=\"italics\">M<\/em> solution is allowed to evaporate until the volume is reduced to 0.105 L?<\/li>\n<li>What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?\n<ol>\n<li>1.00 L of a 0.250-<em data-effect=\"italics\">M<\/em> solution of Fe(NO<sub>3<\/sub>)<sub>3<\/sub> is diluted to a final volume of 2.00 L<\/li>\n<li>0.5000 L of a 0.1222-<em data-effect=\"italics\">M<\/em> solution of C<sub>3<\/sub>H<sub>7<\/sub>OH is diluted to a final volume of 1.250 L<\/li>\n<li>2.35 L of a 0.350-<em data-effect=\"italics\">M<\/em> solution of H<sub>3<\/sub>PO<sub>4<\/sub> is diluted to a final volume of 4.00 L<\/li>\n<li>22.50 mL of a 0.025-<em data-effect=\"italics\">M<\/em> solution of C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub> is diluted to 100.0 mL<\/li>\n<\/ol>\n<\/li>\n<li>What is the final concentration of the solution produced when 225.5 mL of a 0.09988-<em data-effect=\"italics\">M<\/em> solution of Na<sub>2<\/sub>CO<sub>3<\/sub> is allowed to evaporate until the solution volume is reduced to 45.00 mL?<\/li>\n<li>A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?<\/li>\n<li>An experiment in a general chemistry laboratory calls for a 2.00-<em data-effect=\"italics\">M<\/em> solution of HCl. How many mL of 11.9 <em data-effect=\"italics\">M<\/em> HCl would be required to make 250 mL of 2.00 <em data-effect=\"italics\">M<\/em> HCl?<\/li>\n<li>What volume of a 0.20-<em data-effect=\"italics\">M<\/em> K<sub>2<\/sub>SO<sub>4<\/sub> solution contains 57 g of K<sub>2<\/sub>SO<sub>4<\/sub>?<\/li>\n<li>The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg\/L. If an industry is discharging hexavalent chromium as potassium dichromate (K<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub>), what is the maximum permissible molarity of that substance?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\n<div class=\"entry-content\">\n<div class=\"im_section\">\n<div class=\"im_section\">\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\n<div class=\"bcc-box bcc-info\">\n<h4>Selected Answers<\/h4>\n<p>2.\u00a0We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.<\/p>\n<p>4. (a) [latex]\\frac{0.444\\text{mol}}{0.654\\text{L}}=0.679{\\text{mol L}}^{-1}=0.679M\\text{;}[\/latex]<\/p>\n<div id=\"fs-idm98103760\" data-type=\"exercise\">\n<div id=\"fs-idm10494144\" data-type=\"solution\">\n<p>(b) First convert mass in grams to moles, and then substitute the proper terms into the definition.<\/p>\n<p>Molar mass of H<sub>3<\/sub>PO<sub>4<\/sub> = 97.995 g\/mol<\/p>\n<p>[latex]{\\text{mol (H}}_{3}{\\text{PO}}_{4}\\text{)}=98.0\\text{g}\\times \\frac{1\\text{mol}}{97.995\\text{g}}=1.00\\text{mol}[\/latex]<\/p>\n<div data-type=\"newline\">[latex]M=\\frac{1.00\\text{mol}}{1.00\\text{L}}=1.00M\\text{;}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(c) Molar mass [Ca(OH)<sub>2<\/sub>] = 79.09 g\/mol<\/div>\n<div data-type=\"newline\">[latex]0.2074\\cancel{\\text{g}}\\times \\frac{\\text{1 mol}}{74.09\\cancel{\\text{g}}}=0.002799\\text{mol}{\\text{Ca(OH)}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\">[latex]\\frac{0.002799\\text{mol}}{0.0400\\text{L}}=0.06998{\\text{mol L}}^{-1}=0.06998M[\/latex] ;<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(d) Molar mass (Na<sub>2<\/sub>SO<sub>4<\/sub>\u202210H<sub>2<\/sub>O) = 322.20 g\/mol<\/div>\n<div data-type=\"newline\">[latex]10,500\\times \\frac{1\\text{mol}}{322.20\\text{g}}=32.6\\text{mol}[\/latex]<\/div>\n<div data-type=\"newline\">[latex]\\frac{32.6\\text{mol}}{18.60\\text{L}}=1.75M\\text{;}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(e) [latex]M=\\frac{\\text{millimoles solute}}{\\text{volume of solution in milliliters}}[\/latex]<\/div>\n<div data-type=\"newline\">[latex]\\frac{{\\text{7.00 mmol I}}_{2}}{\\text{100 mL}}=0.070M\\text{;}[\/latex] (f) Molar mass (HCl) = 36.46 g\/mol<\/div>\n<div data-type=\"newline\">[latex]\\text{mass (HCl)}=1.8\\times {10}^{1}\\text{g HCl}\\times \\frac{\\text{1 mol}}{\\text{36.46 g}}=\\text{0.49 mol HCl}[\/latex]<\/div>\n<div data-type=\"newline\">[latex]\\frac{\\text{0.49 mol HCl}}{\\text{0.075 L}}=\\text{6.6}M[\/latex]<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\"><\/div>\n<div id=\"fs-idm78757216\" data-type=\"exercise\">6. \u00a0(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass;<\/div>\n<div data-type=\"exercise\"><\/div>\n<div data-type=\"exercise\">(b) 0.500 L contains [latex]0.30M\\times 0.500\\text{L}=1.5\\times {10}^{-1}\\text{mol.}[\/latex] Molar mass (glucose): [latex]6\\times \\text{12.0011 g}+12\\times \\text{1.00794 g}+6\\times \\text{15.9994 g}=\\text{180.158 g},1.5\\times {10}^{-1}\\cancel{\\text{mol}}\\times \\text{180.158 g\/}\\cancel{\\text{mol}}=\\text{27 g.}[\/latex]<\/div>\n<div data-type=\"exercise\"><\/div>\n<div data-type=\"exercise\">8. The molarity must be converted to moles of solute, which is then converted to grams of solute:<span id=\"fs-idm106256160\" data-type=\"media\" data-alt=\"Three boxes connected by right-facing arrows in between each are shown. Written inside the boxes are the phrases, \u201cVolume of solution,\u201d \u201cmoles of solute,\u201d and \u201cmass of solute,\u201d respectively from left to right.\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211206\/CNX_Chem_03_04_Ex0408_img1.jpg\" alt=\"Three boxes connected by right-facing arrows in between each are shown. Written inside the boxes are the phrases, \u201cVolume of solution,\u201d \u201cmoles of solute,\u201d and \u201cmass of solute,\u201d respectively from left to right.\" width=\"884\" height=\"130\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idm107786288\">[latex]M=\\frac{\\text{mol}}{\\text{liter}}\\text{or mol}=M\\times \\text{liter}[\/latex]<\/p>\n<div data-type=\"newline\">(a) [latex]\\begin{array}{l}\\\\ {\\text{mol H}}_{2}{\\text{SO}}_{4}=2.00\\cancel{\\text{L}}\\times \\frac{18.5\\text{mol}}{\\cancel{\\text{L}}}=37.0\\text{mol}{\\text{H}}_{2}{\\text{SO}}_{4}\\\\ 37.0\\cancel{{\\text{mol H}}_{2}{\\text{SO}}_{4}}\\times \\frac{98.08{\\text{g H}}_{2}{\\text{SO}}_{4}}{1\\cancel{{\\text{mol H}}_{2}{\\text{SO}}_{4}}}=3.63\\times {10}^{3}{\\text{g H}}_{2}{\\text{SO}}_{4}\\end{array}\\text{;}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(b) [latex]\\begin{array}{l}\\\\ \\text{mol NaCN}=0.1000\\cancel{\\text{L}}\\times \\frac{3.8\\times {10}^{-5}\\text{mol}}{\\cancel{\\text{L}}}=3.8\\times {10}^{-6}\\text{mol NaCN}\\\\ 3.8\\times {10}^{-5}\\cancel{\\text{mol NaCN}}\\times \\frac{49.01\\text{g}}{1\\cancel{\\text{mol NaCN}}}=1.9\\times {\\text{10}}^{-4}\\text{g NaCN}\\end{array}\\text{;}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(c)[latex]\\begin{array}{l}\\\\ {\\text{mol H}}_{2}\\text{CO}=5.50\\cancel{\\text{L}}\\times \\frac{13.3\\text{mol}}{\\cancel{\\text{L}}}=73.2\\text{mol}{\\text{H}}_{2}\\text{CO}\\\\ 73.2\\cancel{\\text{mol}{\\text{H}}_{2}\\text{CO}}\\times \\frac{30.026\\text{g}}{1\\cancel{{\\text{mol H}}_{2}\\text{CO}}}=2198\\text{g}{\\text{H}}_{2}\\text{CO}=2.20\\text{kg}{\\text{H}}_{2}\\text{CO}\\end{array}\\text{;}[\/latex]<\/div>\n<p>(d) [latex]\\begin{array}{l}\\\\ {\\text{mol FeSO}}_{4}=0.325\\cancel{\\text{L}}\\times \\frac{1.8\\times {10}^{-6}\\text{mol}}{\\cancel{L}}=5.9\\times {10}^{-7}{\\text{mol FeSO}}_{4}\\\\ 5.85\\times {10}^{-7}\\cancel{{\\text{mol FeSO}}_{4}}\\times \\frac{151.9\\text{g}}{1\\cancel{{\\text{mol FeSO}}_{4}}}=8.9\\times {10}^{-5}{\\text{g FeSO}}_{4}\\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>10. (a) Determine the molar mass of KMnO<sub>4<\/sub>; determine the number of moles of KMnO<sub>4<\/sub> in the solution; from the number of moles and the volume of solution, determine the molarity;<\/p>\n<p>(b) Molar mass of KMnO<sub>4<\/sub> = 158.0264 g\/mol<\/p>\n<p>[latex]\\begin{array}{l}\\\\ \\text{mol}{\\text{KMnO}}_{4}=0.0908\\cancel{\\text{g}{\\text{KMnO}}_{4}}\\times \\frac{\\text{1 mol}}{158.0264\\cancel{{\\text{g KMnO}}_{4}}}=5.746\\times {10}^{-4}\\text{mol}\\\\ M{\\text{KMnO}}_{4}=\\frac{5.746\\times {10}^{-4}\\text{mol}}{0.500\\text{L}}=1.15\\times {10}^{-3}M\\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>12. (a) [latex]M{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}=\\frac{\\text{mol}}{V}=\\frac{\\frac{0.195\\cancel{g}{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}}{386.660\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}}}{0.100\\text{L}}=5.04\\times {10}^{-3}M\\text{;}[\/latex]<\/p>\n<p>(b) [latex]M{\\text{NH}}_{3}=\\frac{\\text{mol}}{V}=\\frac{\\frac{4.25\\cancel{g}{\\text{NH}}_{3}}{17.0304\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{NH}}_{3}}}{0.500\\text{L}}=0.499M\\text{;}[\/latex]<\/p>\n<p>(c) [latex]M{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}=\\frac{\\text{mol}}{V}=\\frac{1.49\\cancel{\\text{kg}}{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}\\times \\frac{1000\\cancel{g}}{1\\cancel{\\text{kg}}}\\times \\frac{1\\text{mol}{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}}{60.096\\cancel{\\text{g}}}}{2.50\\text{L}}=9.92M\\text{;}[\/latex]<\/p>\n<p>(d) [latex]M{\\text{I}}_{2}=\\frac{\\text{mol}}{V}=\\frac{\\frac{0.029\\cancel{\\text{g}}{\\text{I}}_{2}}{253.8090\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{I}}_{2}}}{0.100\\text{L}}=1.1\\times {10}^{-3}M[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>14.\u00a0[latex]M=\\frac{\\text{mol}}{V}=\\frac{\\frac{1.0\\cancel{\\text{g}}}{40.08\\cancel{\\text{g}}{\\text{mol}}^{-1}}}{1.0\\text{L}}=0.025M[\/latex]<\/p>\n<p>16.\u00a0[latex]\\begin{array}{c}\\frac{{C}_{1}{V}_{1}}{{C}_{2}}={V}_{2}\\\\ \\frac{\\frac{0.3556\\text{mol}}{\\text{L}}\\times 0.1718\\text{L}}{\\frac{0.1222\\text{mol}}{\\text{L}}}={V}_{2}\\\\ 0.5000\\text{L}={V}_{2}\\end{array}[\/latex]<\/p>\n<p>18.\u00a0[latex]{V}_{1}=\\frac{{V}_{2}\\times {M}_{2}}{{M}_{2}}=25\\text{mL}\\times \\frac{0.025M}{0.33M}=1.9\\text{mL}[\/latex]<\/p>\n<p>20. (a) [latex]{C}_{2}=\\frac{{V}_{1}\\times {C}_{1}}{{V}_{2}}=1.00\\cancel{\\text{L}}\\times \\frac{0.250M}{2.00\\cancel{\\text{L}}}=0.125M\\text{;}[\/latex]<\/p>\n<div data-type=\"exercise\">(b) [latex]{C}_{2}=\\frac{{V}_{1}\\times {C}_{1}}{{V}_{2}}=0.5000\\cancel{\\text{L}}\\times \\frac{0.1222M}{1.250\\cancel{\\text{L}}}=0.04888M\\text{;}[\/latex]<\/div>\n<div data-type=\"exercise\"><\/div>\n<div data-type=\"exercise\">(c) [latex]{C}_{2}=\\frac{{V}_{1}\\times {C}_{1}}{{V}_{2}}=2.35\\cancel{\\text{L}}\\times \\frac{0.350M}{4.00\\cancel{\\text{L}}}=0.206M\\text{;}[\/latex]<\/div>\n<div data-type=\"exercise\"><\/div>\n<div data-type=\"exercise\">(d) [latex]{C}_{2}=\\frac{{V}_{1}\\times {C}_{1}}{{V}_{2}}=0.02250\\cancel{\\text{mL}}\\times \\frac{0.025M}{0.100\\cancel{\\text{mL}}}=0.0056M[\/latex]<\/div>\n<div data-type=\"exercise\"><\/div>\n<div data-type=\"exercise\">22. Determine the number of moles in 434.4 g of HCl: 1.00794 + 35.4527 = 36.4606 g\/mol<\/div>\n<div data-type=\"exercise\">\n<div id=\"fs-idm27373392\" data-type=\"exercise\">\n<div id=\"fs-idm27372432\" data-type=\"solution\">\n<div data-type=\"newline\">[latex]\\text{mol HCl}=\\frac{434.4\\cancel{g}}{36.4606\\cancel{g}{\\text{mol}}^{-1}}=11.91\\text{mol}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">This HCl is present in 1.00 L, so the molarity is 11.9 <em data-effect=\"italics\">M<\/em>.<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm27364240\" data-type=\"exercise\">\n<p>&nbsp;<\/p>\n<p>24.\u00a0[latex]57\\text{g}{\\text{K}}_{2}{\\text{SO}}_{4}\\times \\frac{1\\text{mol}}{174.26\\text{g}}\\times \\frac{1\\text{L}}{0.20\\text{mol}}=1.6\\text{L}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-success\">\n<section id=\"glossary\">\n<h3>Glossary<\/h3>\n<div data-type=\"definition\">\n<div id=\"fs-idm8143856\" data-type=\"definition\">\n<p data-type=\"definition\"><strong><span data-type=\"term\">aqueous solution<br \/>\n<\/span><\/strong>solution for which water is the solvent<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">concentrated<br \/>\n<\/span><\/strong>qualitative term for a solution containing solute at a relatively high concentration<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">concentration<br \/>\n<\/span><\/strong>quantitative measure of the relative amounts of solute and solvent present in a solution<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">dilute<br \/>\n<\/span><\/strong>qualitative term for a solution containing solute at a relatively low concentration<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">dilution<br \/>\n<\/span><\/strong>process of adding solvent to a solution in order to lower the concentration of solutes<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">dissolved<br \/>\n<\/span><\/strong>describes the process by which solute components are dispersed in a solvent<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">molarity (<em data-effect=\"italics\">M<\/em>)<br \/>\n<\/span><\/strong>unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution<\/p>\n<p data-type=\"definition\"><strong><span data-type=\"term\">solute<br \/>\n<\/span><\/strong>solution component present in a concentration less than that of the solvent<\/p>\n<div id=\"fs-idm26666032\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">solvent<br \/>\n<\/span><\/strong>solution component present in a concentration that is higher relative to other components<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1656\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":5,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1656","chapter","type-chapter","status-publish","hentry"],"part":3031,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/1656","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/users\/5"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/1656\/revisions"}],"predecessor-version":[{"id":5113,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/1656\/revisions\/5113"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/parts\/3031"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/1656\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/media?parent=1656"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapter-type?post=1656"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/contributor?post=1656"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/license?post=1656"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}