{"id":2916,"date":"2016-08-24T19:46:53","date_gmt":"2016-08-24T19:46:53","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2916"},"modified":"2017-08-31T17:54:15","modified_gmt":"2017-08-31T17:54:15","slug":"freezing-point-depression","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/freezing-point-depression\/","title":{"raw":"Freezing Point Depression","rendered":"Freezing Point Depression"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-OTMzNTFiMmU1YzQ3ODhkMDVjOWE4MzQ0NTQ1OThmNzQ.-ubj\">\r\n \t<li>Define freezing point depression.<\/li>\r\n \t<li>Calculate the freezing point of a solution when given the molal freezing-point depression constant.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Why salt icy roads?<\/h3>\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212453\/20140811155609133194.jpeg\" alt=\"Salting roads lowers the freezing point, which will cause ice to melt more quickly\" width=\"400\" height=\"273\" \/> Salt truck. Image from <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Salt_truck_Milwaukee.jpg\">Wikimedia<\/a>.[\/caption]\r\n<p id=\"x-ck12-NDg4YzVkYjQyN2EyOGIzYjIxNmEwMDAxYzM5Y2Q1YTg.-foi\">Colligative properties have practical applications, such as the salting of roads in cold-weather climates. By applying salt to an icy road, the melting point of the ice is decreased, and the ice will melt more quickly, making driving safer. Sodium chloride (NaCl) and either calcium chloride (CaCl <sub> 2 <\/sub> ) or magnesium chloride (MgCl <sub> 2 <\/sub> ) are used most frequently, either alone or in a mixture. Sodium chloride is the least expensive option, but is less effective because it only dissociates into two ions instead of three.<\/p>\r\n\r\n<\/div>\r\n<h2>Freezing Point Depression<\/h2>\r\n<p id=\"x-ck12-OTlhZTg5NzZjMmRiNDM0NzU4ZDYyYTg4YWU2ZjY4M2E.-ajp\">The figure<strong>\u00a0<\/strong>below shows the phase diagram for a pure solvent and how it changes when a solute is added to it. The solute lowers the vapor pressure of the solvent resulting in a lowering of the freezing point of the solution compared to the solvent. The <strong> freezing point depression <\/strong> is the difference in temperature between the freezing point of the pure solvent and that of the solution. On the graph, the freezing point depression is represented by <img id=\"x-ck12-MTM2NjcwOTI5MDE0Ng..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212454\/613662958119edecbf2d1a35cf5fc4cd.png\" alt=\"Delta T_f\" width=\"32\" height=\"18\" \/> .<\/p>\r\n\r\n<div id=\"x-ck12-MGMyYTQyMDJlNTNlNjJjZGU5NmVlYzU5NTE2ZWNkNWQ.-awx\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img id=\"x-ck12-OTgwNDUtMTM2Mzc3NzY4Mi04NC0xNC01OA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212454\/20140811155609280023.png\" alt=\"The presence of solute causes a decrease in freezing point by decreasing vapor pressure\" width=\"500\" height=\"420\" longdesc=\"The%20vapor%20pressure%20of%20a%20solution%20%28blue%29%20is%20lower%20than%20the%20vapor%20pressure%20of%20a%20pure%20solvent%20%28pink%29.%20As%20a%20result%2C%20the%20freezing%20point%20of%20a%20solvent%20decreases%20when%20any%20solute%20is%20dissolved%20into%20it.\" \/> Figure 1. The vapor pressure of a solution (blue) is lower than the vapor pressure of a pure solvent (pink). As a result, the freezing point of a solvent decreases when any solute is dissolved into it. Figure from CK-12 Foundation - Christopher Auyeung.[\/caption]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"x-ck12-MmQ3YzM3YTQ2M2RjMDRmOTE1YzdiNGI4Mzc0NzE5ZmU.-5lv\">When a pure solvent freezes, its particles become more ordered as the intermolecular forces that operate between the molecules become permanent. In the case of water, the hydrogen bonds make the hexagonally-shaped network of molecules that characterizes the structure of ice. By dissolving a solute into the liquid solvent, this ordering process is disrupted. As a result, more energy must be removed from the solution in order to freeze it, and the freezing point of the solution is lower than that of the pure solvent.<\/p>\r\n<p id=\"x-ck12-MmFmMWJmZGVjZjg3YjhiN2VlNDc4NGEzMmE1ZGM3MmU.-rjf\">The magnitude of the freezing point depression is directly proportional to the molality of the solution. The equation is:<\/p>\r\n<p id=\"x-ck12-7js\"><img class=\"x-ck12-block-math aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/3c049a7b340e97e90d8c4573d8f7baea.png\" alt=\"Delta T_f=K_f times m\" width=\"119\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-YjAxZTU1MWU2M2EyZGI5YjdmNmE2MDExMmRmMTE4YzU.-mc0\">The proportionality constant, <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/d358324009cfdaad1a755c105a2819f8.png\" alt=\"K_f\" width=\"22\" height=\"18\" \/> , is called the <strong> molal freezing-point depression constant <\/strong> . It is a constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of\u00a0 <img id=\"x-ck12-MTM2NjcwOTI5MDE0Nw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/d358324009cfdaad1a755c105a2819f8.png\" alt=\"K_f\" width=\"22\" height=\"18\" \/> is -1.86\u00b0C\/ <em> m <\/em> . So the freezing temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is -1.86\u00b0C. Every solvent has a unique molal freezing-point depression constant. These are shown in <strong> Table <\/strong> <a href=\"#x-ck12-NjQxN2RmMzExNzc4ZjliODM2ZGNiODFmMjMyNTdlODA.-dhw\"> below <\/a> , along with a related value for the boiling point called <sub><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> . <\/sub><\/p>\r\n\r\n<table id=\"x-ck12-NjQxN2RmMzExNzc4ZjliODM2ZGNiODFmMjMyNTdlODA.-dhw\" class=\"x-ck12-nofloat\" border=\"1\"><caption>Molal Freezing-Point and Boiling-Point Constants<\/caption>\r\n<tbody>\r\n<tr>\r\n<td><strong> Solvent <\/strong><\/td>\r\n<td><strong> Normal freezing point (\u00b0C) <\/strong><\/td>\r\n<td><strong> Molal freezing-point depression constant, K <sub> f <\/sub> (\u00b0C\/ <em> m <\/em> ) <\/strong><\/td>\r\n<td><strong> Normal boiling point (\u00b0C) <\/strong><\/td>\r\n<td><strong> Molal boiling-point elevation constant, Kb (\u00b0C\/ <em> m <\/em> ) <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Acetic acid<\/td>\r\n<td>16.6<\/td>\r\n<td>-3.90<\/td>\r\n<td>117.9<\/td>\r\n<td>3.07<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Camphor<\/td>\r\n<td>178.8<\/td>\r\n<td>-39.7<\/td>\r\n<td>207.4<\/td>\r\n<td>5.61<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Naphthalene<\/td>\r\n<td>80.2<\/td>\r\n<td>-6.94<\/td>\r\n<td>217.7<\/td>\r\n<td>5.80<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Phenol<\/td>\r\n<td>40.9<\/td>\r\n<td>-7.40<\/td>\r\n<td>181.8<\/td>\r\n<td>3.60<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Water<\/td>\r\n<td>0.00<\/td>\r\n<td>-1.86<\/td>\r\n<td>100.00<\/td>\r\n<td>0.512<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h3>Sample Problem: Freezing Point of a Nonelectrolyte<\/h3>\r\n<div id=\"x-ck12-ZGFlZDk0ZTkxYTIxMTY0ZTk5ODcwMTVhOWQ3M2NiNWY.-spu\">\r\n<p id=\"x-ck12-ZGFlZDk0ZTkxYTIxMTY0ZTk5ODcwMTVhOWQ3M2NiNWY.-ztx\">Ethylene glycol (C <sub> 2 <\/sub> H <sub> 6 <\/sub> O <sub> 2 <\/sub> ) is a molecular compound that is used in many commercial anti-freezes. A water solution of ethylene glycol is used in vehicle radiators to lower its freezing point and thus prevent the water in the radiator from freezing. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500. g of water.<\/p>\r\n\r\n<\/div>\r\n<div id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-rhz\">\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-kz6\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-erf\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MzJmNjExODYzM2ZmM2U3Y2Y5MWIxOGMxNDE1ZDdmOTE.-fzm\">\r\n \t<li>mass C <sub> 2 <\/sub> H <sub> 6 <\/sub> O <sub> 2 <\/sub> = 400. g<\/li>\r\n \t<li>molar mass C <sub> 2 <\/sub> H <sub> 6 <\/sub> O <sub> 2 <\/sub> = 62.08 g\/mol<\/li>\r\n \t<li>mass H <sub> 2 <\/sub> O = 500.0 g = 0.500 kg<\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/a2d2b094697683867a86403f98ea6369.png\" alt=\"text{mass H}_2text{O}=500.0 text{g}=0.500 text{kg}\" width=\"247\" height=\"16\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212457\/1b4605e56fb75eb041a94739afcb9ee8.png\" alt=\"K_f(text{H}_2text{O})=-1.86^circ text{C}\/m\" width=\"187\" height=\"20\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-9kt\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-u6n\">\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212458\/e2e3d6869b877f52936da7cfd2cffc50.png\" alt=\"T_f text{of solution}=? ^circ text{C}\" width=\"160\" height=\"19\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-M2Q3Yzg0YWJkNDQ3ZDllYjllNGMzMDk1MGE1MThmOTg.-obg\">This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression.<\/p>\r\n\r\n<\/div>\r\n<div id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-ls7\">\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-7ou\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-nxz\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212459\/25968e351ded7e25901b5e7c71796ea3.png\" alt=\"400. text{ g C}_2text{H}_6text{O}_2 times frac{1 text{ mol C}_2text{H}_6text{O}_2}{62.08 text{ g C}_2text{H}_6text{O}_2} &amp;= 6.44 text{ mol C}_2text{H}_6text{O}_2\\frac{6.44 text{ mol C}_2text{H}_6text{O}_2}{0.500 text{ kg H}_2text{O}} &amp;= 12.9 m text{C}_2text{H}_6text{O}_2\\Delta T_f=K_f times m=-1.86^circ text{C\/}m times 12.9 m &amp;= -24.0^circ text{C} \\T_f &amp;=-24.0^circ text{C}\" width=\"467\" height=\"144\" \/><\/p>\r\n\r\n<\/div>\r\n<div id=\"x-ck12-ZjdlYWIyMjk4OThhOWMyYzBkZWViNzJhMDBjOTE3MGU.-qio\">\r\n<p id=\"x-ck12-ZjdlYWIyMjk4OThhOWMyYzBkZWViNzJhMDBjOTE3MGU.-a8g\">The normal freezing point of water is 0.0\u00b0C. Therefore, since the freezing point decreases by 24.0\u00b0C, the freezing point of the solution is -24.0\u00b0C.<\/p>\r\n\r\n<\/div>\r\n<div id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-sc2\">\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-ar3\"><em> Step 3: Think about your result. <\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"x-ck12-OTE2ZWI3YmM5NmQ4MWNiZmZlZjhiMTA1ODJkZjA5M2Y.-a0u\">\r\n<p id=\"x-ck12-OTE2ZWI3YmM5NmQ4MWNiZmZlZjhiMTA1ODJkZjA5M2Y.-kii\">The freezing point of the water decreases by a large amount, protecting the radiator from damage due to the expansion of water when it freezes. There are three significant figures in the result.<\/p>\r\n\r\n<\/div>\r\n<div id=\"x-ck12-MjkwNjEyMTk5ODYxYzMxZDEwMzZiMTg1YjRlNjliNzU.-gbj\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<div id=\"x-ck12-MjkwNjEyMTk5ODYxYzMxZDEwMzZiMTg1YjRlNjliNzU.-gbj\"><\/div>\r\n<div id=\"x-ck12-YzBmYmRjMTQxZDAwMDkxMjY2ZTYwZjNkMDlhZDI2MTU.-lbs\">\r\n<ul id=\"x-ck12-OWFmYTg3N2MxMjA4Zjg0YmZlN2I4NGFlZWI3YzA3ZTQ.-lqa\">\r\n \t<li>Freezing point depression is defined.<\/li>\r\n \t<li>Calculations involving freezing point depression are described.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"x-ck12-YzBmYmRjMTQxZDAwMDkxMjY2ZTYwZjNkMDlhZDI2MTU.-lbs\">\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ODA3NjMzNDIzMDE3OWNmYzFhNDViZDZiOGM2ZTVkNTM.-ofy\">\r\n \t<li>How does a solute affect the freezing of water?<\/li>\r\n \t<li>How many moles of glucose would be needed to lower the freezing point of one kg of water 3.72\u00b0C?<\/li>\r\n \t<li>How many moles of NaCl would be needed to produce the same amount of lowering of temperature?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h2 class=\"x-ck12-data-problem-set\">\u00a0Glossary<\/h2>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ODJmNTU1MGY4MDIzOWQwZTY2NjE5ZTc0NzdlMzQ4MDE.-hz6\">\r\n \t<li><strong> freezing point depression: <\/strong> The difference in temperature between the freezing point of the pure solvent and that of the solution.<\/li>\r\n \t<li><strong> molal freezing-point depression constant: <\/strong> A constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute.<\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-OTMzNTFiMmU1YzQ3ODhkMDVjOWE4MzQ0NTQ1OThmNzQ.-ubj\">\n<li>Define freezing point depression.<\/li>\n<li>Calculate the freezing point of a solution when given the molal freezing-point depression constant.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Why salt icy roads?<\/h3>\n<div style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212453\/20140811155609133194.jpeg\" alt=\"Salting roads lowers the freezing point, which will cause ice to melt more quickly\" width=\"400\" height=\"273\" \/><\/p>\n<p class=\"wp-caption-text\">Salt truck. Image from <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Salt_truck_Milwaukee.jpg\">Wikimedia<\/a>.<\/p>\n<\/div>\n<p id=\"x-ck12-NDg4YzVkYjQyN2EyOGIzYjIxNmEwMDAxYzM5Y2Q1YTg.-foi\">Colligative properties have practical applications, such as the salting of roads in cold-weather climates. By applying salt to an icy road, the melting point of the ice is decreased, and the ice will melt more quickly, making driving safer. Sodium chloride (NaCl) and either calcium chloride (CaCl <sub> 2 <\/sub> ) or magnesium chloride (MgCl <sub> 2 <\/sub> ) are used most frequently, either alone or in a mixture. Sodium chloride is the least expensive option, but is less effective because it only dissociates into two ions instead of three.<\/p>\n<\/div>\n<h2>Freezing Point Depression<\/h2>\n<p id=\"x-ck12-OTlhZTg5NzZjMmRiNDM0NzU4ZDYyYTg4YWU2ZjY4M2E.-ajp\">The figure<strong>\u00a0<\/strong>below shows the phase diagram for a pure solvent and how it changes when a solute is added to it. The solute lowers the vapor pressure of the solvent resulting in a lowering of the freezing point of the solution compared to the solvent. The <strong> freezing point depression <\/strong> is the difference in temperature between the freezing point of the pure solvent and that of the solution. On the graph, the freezing point depression is represented by <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjcwOTI5MDE0Ng..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212454\/613662958119edecbf2d1a35cf5fc4cd.png\" alt=\"Delta T_f\" width=\"32\" height=\"18\" \/> .<\/p>\n<div id=\"x-ck12-MGMyYTQyMDJlNTNlNjJjZGU5NmVlYzU5NTE2ZWNkNWQ.-awx\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2Mzc3NzY4Mi04NC0xNC01OA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212454\/20140811155609280023.png\" alt=\"The presence of solute causes a decrease in freezing point by decreasing vapor pressure\" width=\"500\" height=\"420\" longdesc=\"The%20vapor%20pressure%20of%20a%20solution%20%28blue%29%20is%20lower%20than%20the%20vapor%20pressure%20of%20a%20pure%20solvent%20%28pink%29.%20As%20a%20result%2C%20the%20freezing%20point%20of%20a%20solvent%20decreases%20when%20any%20solute%20is%20dissolved%20into%20it.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The vapor pressure of a solution (blue) is lower than the vapor pressure of a pure solvent (pink). As a result, the freezing point of a solvent decreases when any solute is dissolved into it. Figure from CK-12 Foundation &#8211; Christopher Auyeung.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"x-ck12-MmQ3YzM3YTQ2M2RjMDRmOTE1YzdiNGI4Mzc0NzE5ZmU.-5lv\">When a pure solvent freezes, its particles become more ordered as the intermolecular forces that operate between the molecules become permanent. In the case of water, the hydrogen bonds make the hexagonally-shaped network of molecules that characterizes the structure of ice. By dissolving a solute into the liquid solvent, this ordering process is disrupted. As a result, more energy must be removed from the solution in order to freeze it, and the freezing point of the solution is lower than that of the pure solvent.<\/p>\n<p id=\"x-ck12-MmFmMWJmZGVjZjg3YjhiN2VlNDc4NGEzMmE1ZGM3MmU.-rjf\">The magnitude of the freezing point depression is directly proportional to the molality of the solution. The equation is:<\/p>\n<p id=\"x-ck12-7js\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/3c049a7b340e97e90d8c4573d8f7baea.png\" alt=\"Delta T_f=K_f times m\" width=\"119\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-YjAxZTU1MWU2M2EyZGI5YjdmNmE2MDExMmRmMTE4YzU.-mc0\">The proportionality constant, <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/d358324009cfdaad1a755c105a2819f8.png\" alt=\"K_f\" width=\"22\" height=\"18\" \/> , is called the <strong> molal freezing-point depression constant <\/strong> . It is a constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjcwOTI5MDE0Nw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/d358324009cfdaad1a755c105a2819f8.png\" alt=\"K_f\" width=\"22\" height=\"18\" \/> is -1.86\u00b0C\/ <em> m <\/em> . So the freezing temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is -1.86\u00b0C. Every solvent has a unique molal freezing-point depression constant. These are shown in <strong> Table <\/strong> <a href=\"#x-ck12-NjQxN2RmMzExNzc4ZjliODM2ZGNiODFmMjMyNTdlODA.-dhw\"> below <\/a> , along with a related value for the boiling point called <sub><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> . <\/sub><\/p>\n<table id=\"x-ck12-NjQxN2RmMzExNzc4ZjliODM2ZGNiODFmMjMyNTdlODA.-dhw\" class=\"x-ck12-nofloat\">\n<caption>Molal Freezing-Point and Boiling-Point Constants<\/caption>\n<tbody>\n<tr>\n<td><strong> Solvent <\/strong><\/td>\n<td><strong> Normal freezing point (\u00b0C) <\/strong><\/td>\n<td><strong> Molal freezing-point depression constant, K <sub> f <\/sub> (\u00b0C\/ <em> m <\/em> ) <\/strong><\/td>\n<td><strong> Normal boiling point (\u00b0C) <\/strong><\/td>\n<td><strong> Molal boiling-point elevation constant, Kb (\u00b0C\/ <em> m <\/em> ) <\/strong><\/td>\n<\/tr>\n<tr>\n<td>Acetic acid<\/td>\n<td>16.6<\/td>\n<td>-3.90<\/td>\n<td>117.9<\/td>\n<td>3.07<\/td>\n<\/tr>\n<tr>\n<td>Camphor<\/td>\n<td>178.8<\/td>\n<td>-39.7<\/td>\n<td>207.4<\/td>\n<td>5.61<\/td>\n<\/tr>\n<tr>\n<td>Naphthalene<\/td>\n<td>80.2<\/td>\n<td>-6.94<\/td>\n<td>217.7<\/td>\n<td>5.80<\/td>\n<\/tr>\n<tr>\n<td>Phenol<\/td>\n<td>40.9<\/td>\n<td>-7.40<\/td>\n<td>181.8<\/td>\n<td>3.60<\/td>\n<\/tr>\n<tr>\n<td>Water<\/td>\n<td>0.00<\/td>\n<td>-1.86<\/td>\n<td>100.00<\/td>\n<td>0.512<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3>Sample Problem: Freezing Point of a Nonelectrolyte<\/h3>\n<div id=\"x-ck12-ZGFlZDk0ZTkxYTIxMTY0ZTk5ODcwMTVhOWQ3M2NiNWY.-spu\">\n<p id=\"x-ck12-ZGFlZDk0ZTkxYTIxMTY0ZTk5ODcwMTVhOWQ3M2NiNWY.-ztx\">Ethylene glycol (C <sub> 2 <\/sub> H <sub> 6 <\/sub> O <sub> 2 <\/sub> ) is a molecular compound that is used in many commercial anti-freezes. A water solution of ethylene glycol is used in vehicle radiators to lower its freezing point and thus prevent the water in the radiator from freezing. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500. g of water.<\/p>\n<\/div>\n<div id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-rhz\">\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-kz6\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-erf\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-MzJmNjExODYzM2ZmM2U3Y2Y5MWIxOGMxNDE1ZDdmOTE.-fzm\">\n<li>mass C <sub> 2 <\/sub> H <sub> 6 <\/sub> O <sub> 2 <\/sub> = 400. g<\/li>\n<li>molar mass C <sub> 2 <\/sub> H <sub> 6 <\/sub> O <sub> 2 <\/sub> = 62.08 g\/mol<\/li>\n<li>mass H <sub> 2 <\/sub> O = 500.0 g = 0.500 kg<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/a2d2b094697683867a86403f98ea6369.png\" alt=\"text{mass H}_2text{O}=500.0 text{g}=0.500 text{kg}\" width=\"247\" height=\"16\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212457\/1b4605e56fb75eb041a94739afcb9ee8.png\" alt=\"K_f(text{H}_2text{O})=-1.86^circ text{C}\/m\" width=\"187\" height=\"20\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-9kt\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-u6n\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212458\/e2e3d6869b877f52936da7cfd2cffc50.png\" alt=\"T_f text{of solution}=? ^circ text{C}\" width=\"160\" height=\"19\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-M2Q3Yzg0YWJkNDQ3ZDllYjllNGMzMDk1MGE1MThmOTg.-obg\">This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression.<\/p>\n<\/div>\n<div id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-ls7\">\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-7ou\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-nxz\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212459\/25968e351ded7e25901b5e7c71796ea3.png\" alt=\"400. text{ g C}_2text{H}_6text{O}_2 times frac{1 text{ mol C}_2text{H}_6text{O}_2}{62.08 text{ g C}_2text{H}_6text{O}_2} &amp;= 6.44 text{ mol C}_2text{H}_6text{O}_2\\frac{6.44 text{ mol C}_2text{H}_6text{O}_2}{0.500 text{ kg H}_2text{O}} &amp;= 12.9 m text{C}_2text{H}_6text{O}_2\\Delta T_f=K_f times m=-1.86^circ text{C\/}m times 12.9 m &amp;= -24.0^circ text{C} \\T_f &amp;=-24.0^circ text{C}\" width=\"467\" height=\"144\" \/><\/p>\n<\/div>\n<div id=\"x-ck12-ZjdlYWIyMjk4OThhOWMyYzBkZWViNzJhMDBjOTE3MGU.-qio\">\n<p id=\"x-ck12-ZjdlYWIyMjk4OThhOWMyYzBkZWViNzJhMDBjOTE3MGU.-a8g\">The normal freezing point of water is 0.0\u00b0C. Therefore, since the freezing point decreases by 24.0\u00b0C, the freezing point of the solution is -24.0\u00b0C.<\/p>\n<\/div>\n<div id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-sc2\">\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-ar3\"><em> Step 3: Think about your result. <\/em><\/p>\n<\/div>\n<div id=\"x-ck12-OTE2ZWI3YmM5NmQ4MWNiZmZlZjhiMTA1ODJkZjA5M2Y.-a0u\">\n<p id=\"x-ck12-OTE2ZWI3YmM5NmQ4MWNiZmZlZjhiMTA1ODJkZjA5M2Y.-kii\">The freezing point of the water decreases by a large amount, protecting the radiator from damage due to the expansion of water when it freezes. There are three significant figures in the result.<\/p>\n<\/div>\n<div id=\"x-ck12-MjkwNjEyMTk5ODYxYzMxZDEwMzZiMTg1YjRlNjliNzU.-gbj\">\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<div id=\"x-ck12-MjkwNjEyMTk5ODYxYzMxZDEwMzZiMTg1YjRlNjliNzU.-gbj\"><\/div>\n<div id=\"x-ck12-YzBmYmRjMTQxZDAwMDkxMjY2ZTYwZjNkMDlhZDI2MTU.-lbs\">\n<ul id=\"x-ck12-OWFmYTg3N2MxMjA4Zjg0YmZlN2I4NGFlZWI3YzA3ZTQ.-lqa\">\n<li>Freezing point depression is defined.<\/li>\n<li>Calculations involving freezing point depression are described.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"x-ck12-YzBmYmRjMTQxZDAwMDkxMjY2ZTYwZjNkMDlhZDI2MTU.-lbs\">\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ODA3NjMzNDIzMDE3OWNmYzFhNDViZDZiOGM2ZTVkNTM.-ofy\">\n<li>How does a solute affect the freezing of water?<\/li>\n<li>How many moles of glucose would be needed to lower the freezing point of one kg of water 3.72\u00b0C?<\/li>\n<li>How many moles of NaCl would be needed to produce the same amount of lowering of temperature?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h2 class=\"x-ck12-data-problem-set\">\u00a0Glossary<\/h2>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ODJmNTU1MGY4MDIzOWQwZTY2NjE5ZTc0NzdlMzQ4MDE.-hz6\">\n<li><strong> freezing point depression: <\/strong> The difference in temperature between the freezing point of the pure solvent and that of the solution.<\/li>\n<li><strong> molal freezing-point depression constant: <\/strong> A constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute.<\/li>\n<\/ul>\n<\/div>\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-2916\">\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 Concepts Intermediate . <strong>Authored by<\/strong>: Calbreath, Baxter, et al.. <strong>Provided by<\/strong>: CK12.org. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em><\/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":29,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate \",\"author\":\"Calbreath, Baxter, et al.\",\"organization\":\"CK12.org\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2916","chapter","type-chapter","status-publish","hentry"],"part":2337,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2916","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/29"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2916\/revisions"}],"predecessor-version":[{"id":3687,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2916\/revisions\/3687"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2337"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2916\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2916"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2916"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2916"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2916"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}