Exercises

Part 1

Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.
What information do we need to calculate the molarity of a sulfuric acid solution?
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
Determine the molarity of each of the following solutions:
1. 1.457 mol KCl in 1.500 L of solution
2. 0.515 g of H₂SO₄ in 1.00 L of solution
3. 20.54 g of Al(NO₃)₃ in 1575 mL of solution
4. 2.76 kg of CuSO₄•5H₂O in 1.45 L of solution
5. 0.005653 mol of Br₂ in 10.00 mL of solution
6. 0.000889 g of glycine, C₂H₅NO₂, in 1.05 mL of solution
Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, C₆H₁₂O₆, used for intravenous injection?
1. Outline the steps necessary to answer the question.
2. Answer the question.
Consider this question: What is the mass of solute in 200.0 L of a 1.556-M solution of KBr?
1. Outline the steps necessary to answer the question.
2. Answer the question.
Calculate the number of moles and the mass of the solute in each of the following solutions:
1. 2.00 L of 18.5 M H₂SO₄, concentrated sulfuric acid
2. 100.0 mL of [latex]3.8\times {10}^{-5}M\text{NaCN,}[/latex] the minimum lethal concentration of sodium cyanide in blood serum
3. 5.50 L of 13.3 M H₂CO, the formaldehyde used to “fix” tissue samples
4. 325 mL of [latex]1.8\times {10}^{-6}M[/latex] FeSO₄, the minimum concentration of iron sulfate detectable by taste in drinking water
Calculate the number of moles and the mass of the solute in each of the following solutions:
1. 325 mL of [latex]8.23\times {10}^{-5}M\text{KI}[/latex], a source of iodine in the diet
2. 75.0 mL of [latex]2.2\times {10}^{-5}M{\text{H}}_{2}{\text{SO}}_{4}[/latex], a sample of acid rain
3. 0.2500 L of 0.1135 M K₂CrO₄, an analytical reagent used in iron assays
4. 10.5 L of 3.716 M (NH₄)₂SO₄, a liquid fertilizer
Consider this question: What is the molarity of KMnO₄ in a solution of 0.0908 g of KMnO₄ in 0.500 L of solution?
1. Outline the steps necessary to answer the question.
2. Answer the question.
Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl?
1. Outline the steps necessary to answer the question.
2. Answer the question.
Calculate the molarity of each of the following solutions:
1. 0.195 g of cholesterol, C₂₇H₄₆O, in 0.100 L of serum, the average concentration of cholesterol in human serum
2. 4.25 g of NH₃ in 0.500 L of solution, the concentration of NH₃ in household ammonia
3. 1.49 kg of isopropyl alcohol, C₃H₇OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol
4. 0.029 g of I₂ in 0.100 L of solution, the solubility of I₂ in water at 20 °C
Calculate the molarity of each of the following solutions:
1. 293 g HCl in 666 mL of solution, a concentrated HCl solution
2. 2.026 g FeCl₃ in 0.1250 L of a solution used as an unknown in general chemistry laboratories
3. 0.001 mg Cd²⁺ in 0.100 L, the maximum permissible concentration of cadmium in drinking water
4. 0.0079 g C₇H₅SNO₃ in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink.
There is about 1.0 g of calcium, as Ca²⁺, in 1.0 L of milk. What is the molarity of Ca²⁺ in milk?
What volume of a 1.00-M Fe(NO₃)₃ solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M?
If 0.1718 L of a 0.3556-M C₃H₇OH solution is diluted to a concentration of 0.1222 M, what is the volume of the resulting solution?
If 4.12 L of a 0.850 M-H₃PO₄ solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution?
What volume of a 0.33-M C₁₂H₂₂O₁₁ solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?
What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M solution is allowed to evaporate until the volume is reduced to 0.105 L?
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?
An experiment in a general chemistry laboratory calls for a 2.00-M solution of HCl. How many mL of 11.9 M HCl would be required to make 250 mL of 2.00 M HCl?
What volume of a 0.20-M K₂SO₄ solution contains 57 g of K₂SO₄?
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 (K₂Cr₂O₇), what is the maximum permissible molarity of that substance?

Show Selected Answers

2. We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.

6. The answers are as follows:

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;
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]

8. The molarity must be converted to moles of solute, which is then converted to grams of solute:
Three boxes connected by right-facing arrows in between each are shown. Written inside the boxes are the phrases, “Volume of solution,” “moles of solute,” and “mass of solute,” respectively from left to right.

[latex]M=\frac{\text{mol}}{\text{liter}}\,\,\,\,\text{or}\,\,\,\,\text{mol}=M\times \text{liter}[/latex]

[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}[/latex]
[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}[/latex]
[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}[/latex]
[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]

10. The answers are as follows:

Determine the molar mass of KMnO₄; determine the number of moles of KMnO₄ in the solution; from the number of moles and the volume of solution, determine the molarity
Molar mass of KMnO₄ = 158.0264 g/mol
[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]

12. The molarity for each solution is as follows:

[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[/latex]
[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[/latex]
[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[/latex]
[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]

14. [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]

16. [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]

18. [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]

22. Determine the number of moles in 434.4 g of HCl: 1.00794 + 35.4527 = 36.4606 g/mol

[latex]\text{mol HCl}=\frac{434.4\cancel{g}}{36.4606\cancel{g}{\text{mol}}^{-1}}=11.91\text{mol}[/latex]

This HCl is present in 1.00 L, so the molarity is 11.9 M.

24. [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]

Part 2

Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO₃ by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO₃ by mass?
1. Outline the steps necessary to answer the question.
2. Answer the question.
What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?
What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm^–3 and contains 37.21% HCl by mass?
The hardness of water (hardness count) is usually expressed in parts per million (by mass) of CaCO₃, which is equivalent to milligrams of CaCO₃ per liter of water. What is the molar concentration of Ca²⁺ions in a water sample with a hardness count of 175 mg CaCO₃/L?
The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g/mL and calculate the molarity of mercury in the stream.
In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM is observed, what is the concentration of glucose (C₆H₁₂O₆) in mg/dL?
A cough syrup contains 5.0% ethyl alcohol, C₂H₅OH, by mass. If the density of the solution is 0.9928 g/mL, determine the molarity of the alcohol in the cough syrup.
D5W is a solution used as an intravenous fluid. It is a 5.0% by mass solution of dextrose (C₆H₁₂O₆) in water. If the density of D5W is 1.029 g/mL, calculate the molarity of dextrose in the solution.
Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, H₂SO₄, for which the density is 1.3057 g/mL.

Show Selected Answers

1. The answers are as follows:

The dilution equation can be used, appropriately modified to accommodate mass-based concentration units:
[latex]\%{\text{mass}}_{1}\times {\text{mass}}_{1}=\%{\text{mass}}_{2}\times {\text{mass}}_{2}[/latex]
This equation can be rearranged to isolate mass₁ and the given quantities substituted into this equation.
[latex]{\text{mass}}_{1}=\frac{\%{\text{mass}}_{2}\times {\text{mass}}_{2}}{\%{\text{mass}}_{1}}=\frac{10.0\%\times 400.0\text{g}}{68.0\%}=58.8\text{g}[/latex]

3. The mass of the 10% solution is [latex]1000{\cancel{\text{cm}}}^{3}\times \frac{1.109\text{g}}{{\cancel{\text{cm}}}^{3}}=1.11\times {10}^{3}\text{g.}[/latex]

The mass of pure NaOH required is [latex]\text{mass}\left(\text{NaOH}\right)=\frac{10.0\%}{100.0\%}\times 1.11\times {10}^{3}\text{g}=1.11\times {10}^{2}\text{g.}[/latex]

This mass of NaOH must come from the 97.0% solution: [latex]\begin{array}{l}\\ \text{mass}\left(\text{NaOH solution}\right)=\frac{97.0\%}{100.0\%}=1.11\times {10}^{2}\text{g}\\ \text{mass}\left(\text{NaOH solution}\right)=\frac{1.11\times {10}^{2}\text{g}}{0.970}=114\text{g}\end{array}[/latex]

5. Since CaCO₃ contains 1 mol Ca²⁺ per mol of CaCO₃, the molar concentration of Ca²⁺ equals the molarity of CaCO₃

[latex]M{\text{Ca}}^{2+}=\frac{\text{mol}{\text{CaCO}}_{3}}{\text{L}}=\frac{175\cancel{\text{mg}}\times \left(\frac{1\text{mol}}{100.0792\cancel{\text{g}}}\right)\times \left(\frac{1\cancel{\text{g}}}{1000\text{mg}}\right)}{1\text{L}}=1.75\times {10}^{-3}M[/latex]

7. 1 mg/dL = 0.01 g/L and 1 L = 10 dL

[latex]5.3\cancel{\text{mmol}}\text{/L}\times 180.158\text{mg/}\cancel{\text{mmol}}=9.5\times {10}^{2}\text{mg/L}[/latex]
[latex]9.5\times {10}^{2}\text{mg/L}\times \frac{1\text{L}}{10\text{dL}}=95\text{mg/dL}[/latex]

11. The molar mass of C₆H₁₂O₆ is [latex]6\times 12.011+12\times 1.00794+6\times 15.9994=180.2\text{g/mol.}[/latex] In 1.000 L, there are:

[latex]\begin{array}{l}\left(1000\cancel{\text{mL}}\times 1.029\text{g}{\cancel{\text{mL}}}^{-1}\right)=1029\text{g}\\ \text{mol dextrose}=1029\cancel{\text{g}}\times 0.050\times \frac{1\text{mol}}{180.2\cancel{\text{g}}}=0.29\text{mol}\end{array}[/latex] C₆H₁₂O₆.

Since we selected the volume to be 1.00 L, the molarity of dextrose is [latex]\text{molarity}=\frac{\text{mol}}{\text{L}}=\frac{0.29\text{mol}}{1.00\text{L}}=0.29\text{mol}[/latex].

Part 3

Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.
Consider the solutions presented:
1. Which of the following sketches best represents the ions in a solution of Fe(NO₃)₃(aq)?
2. Write a balanced chemical equation showing the products of the dissolution of Fe(NO₃)₃.
Compare the processes that occur when methanol (CH₃OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.
What is the expected electrical conductivity of the following solutions?
1. NaOH(aq)
2. HCl(aq)
3. C₆H₁₂O₆(aq) (glucose)
4. NH₃(l)
Why are most solid ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid (molten) ionic compound to be electrically conductive or nonconductive? Explain.
Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:
1. the solutions in Figure 2
2. methanol, CH₃OH, dissolved in ethanol, C₂H₅OH
3. methane, CH₄, dissolved in benzene, C₆H₆
4. the polar halocarbon CF₂Cl₂ dissolved in the polar halocarbon CF₂ClCFCl₂
5. O₂(l) in N₂(l)

Show Selected Answers

Chapter 5-Water

Part 1

Part 3