{"id":2922,"date":"2016-08-24T19:48:12","date_gmt":"2016-08-24T19:48:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2922"},"modified":"2016-08-26T18:57:01","modified_gmt":"2016-08-26T18:57:01","slug":"electrolytes-and-colligative-properties","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/electrolytes-and-colligative-properties\/","title":{"raw":"Electrolytes and Colligative Properties","rendered":"Electrolytes and Colligative Properties"},"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-ODk0MjAzMDJjZDlmZTU2MDg0NjMyZDU0OWRjOTRiODk.-ief\">\r\n \t<li>Perform calculations involving the effects of electrolytes on colligative properties.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>What effect do ions have?<\/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\/19212505\/20140811155609593936.jpeg\" alt=\"The presence of ions in solutions affect the properties of that solution\" width=\"400\" height=\"375\" \/> Ionic solution measurement. Image from <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Ionic_solution_measure.jpg\">Wikimedia<\/a>.[\/caption]\r\n<p id=\"x-ck12-MjNjZjZiOTg2ZjViNzFhMzc1YWRmZTJiYTgxOTM3OGQ.-ghp\">The addition of ions creates significant changes in properties of solutions. Water molecules surround the ions and are somewhat tightly bound to them. Colligative properties are affected because the solvent properties are no longer the same as those in the pure solvent.<\/p>\r\n\r\n<\/div>\r\n<h2>Electrolytes and Colligative Properties<\/h2>\r\n<p id=\"x-ck12-ZGFlMjM2M2I4NjQwOWU5MGZkZjZjMDQ3YzgyNDEzNWU.-brr\">Ionic compounds are electrolytes and dissociate into two or more ions as they dissolve. This must be taken into account when calculating the freezing and boiling points of electrolyte solutions. The sample problem below demonstrates how to calculate the freezing point and boiling point of a solution of calcium chloride. Calcium chloride dissociates into three ions according to the equation:<\/p>\r\n<p id=\"x-ck12-ypv\"><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\/19212507\/85c2e5b28a3943bea0f1cd919f987b18.png\" alt=\"text{CaCl}_2 (s) rightarrow text{Ca}^{2+} (aq)+2text{Cl}^-(aq)\" width=\"259\" height=\"21\" \/><\/p>\r\n<p id=\"x-ck12-ZWRjNWVkMmFiODdmNzEyZjYwYjY4YTY0OTE5MjlmMDE.-yse\">The values of the freezing point depression and the boiling point elevation for a solution of CaCl <sub> 2 <\/sub> will be three times greater than they would be for an equal molality of a nonelectrolyte.<\/p>\r\n\r\n<h3>Sample Problem: Freezing and Boiling Point of an Electrolyte<\/h3>\r\n<p id=\"x-ck12-MDlmZDZiOThjYTZkZDIwNWFkZjU3MWFmOTk2ZTFlMTY.-vad\">Determine the freezing and boiling point of a solution prepared by dissolving 82.20 g of calcium chloride into 400.g of water.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-shk\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-h5d\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NjJiMDRkYWE5MTgzNTQzNTdiMTlkOGIyMjY3ZTY4ZTQ.-so3\">\r\n \t<li>mass CaCl <sub> 2 <\/sub> = 82.20 g<\/li>\r\n \t<li>molar mass CaCl <sub> 2 <\/sub> = 110.98 g\/mol<\/li>\r\n \t<li>mass H <sub> 2 <\/sub> O = 400. g = 0.400 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\/19212508\/122575991dc39025b18d03758f3d1ce4.png\" alt=\"K_f(text{H}_2text{O})=-1.86^circ text{C}\/ m\" width=\"193\" height=\"20\" \/><\/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\/19212509\/58216730de76ccbd7e9e882e3169b515.png\" alt=\"K_b(text{H}_2text{O})=0.512^circ text{C}\/ m\" width=\"186\" height=\"19\" \/><\/li>\r\n \t<li>CaCl <sub> 2 <\/sub> dissociates into 3 ions<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-ljk\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-0m5\">\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\/19212510\/14603dff3c635ec418dafe7bd0063a8f.png\" alt=\"T_f=? ^circ text{C}\" width=\"72\" height=\"18\" \/><\/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\/19212511\/a6045c6e411bb44f334cd9e7d13eae41.png\" alt=\"T_b=? ^circ text{C}\" width=\"70\" height=\"15\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MDY1YjdiOTI3NTFkZWE3MjI4YWVjZGMxMjc3NDE3ZTc.-g0u\">The moles of CaCl <sub> 2 <\/sub> is first calculated, followed by the molality of the solution. The freezing and boiling points are then determined, including multiplying by 3 for the three ions.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-z3o\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-yzq\"><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\/19212511\/994717590bfed24072df301446b06071.png\" alt=\"82.20 text{ g CaCl}_2 times frac{1 text{ mol CaCl}_2}{110.98 text{ g CaCl}_2} &amp;= 0.7407 text{ mol CaCl}_2\\frac{0.7407 text{ mol CaCl}_2}{0.400 text{ kg H}_2text{O}} &amp;= 1.85 m text{CaCl}_2\" width=\"413\" height=\"90\" \/><\/p>\r\n<p id=\"x-ck12-mzv\"><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\/19212514\/b2c96bb7c732a35c895343a1b5695f32.png\" alt=\"Delta T_f &amp;= K_f times m times 3=-1.86^circ text{C}\/m times 1.85 m times 3=-10.3^circ text{C} &amp;&amp; T_f=-10.3^circ text{C}\\Delta T_b &amp;= K_b times m times 3=0.512^circ text{C}\/m times 1.85 m times 3=2.84^circ text{C} &amp;&amp; T_b=102.84^circ text{C}\" width=\"587\" height=\"45\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-nfv\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-Y2M4Mzg0MzYwY2FkY2RjYTVkMGQyZWEzYjVkNWVjMWM.-a9s\">Since the normal boiling point of water is 100.00\u00b0C, the calculated result for <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\/19212501\/0e43da3d9338e5a4c8e4daa908121228.png\" alt=\"Delta T_b\" width=\"31\" height=\"15\" \/> \u00a0must be added to 100.00 to find the new boiling point.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-OWYzOTU0NWU0ZTE2ODNmNWE5ZjY0OTllNjU2NGI3ODg.-wcm\">\r\n \t<li>The effect of ionization on colligative properties is described.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Practice<\/h3>\r\n<p id=\"x-ck12-Y2ZkODlkZGNmOGY3MDBkMzBmYmEzMTdlYTZiODRmNGQ.-ide\">Do the practice and homework problems dealing with ionic solutions toward the end of the section on the link below:<\/p>\r\n<p id=\"x-ck12-YTA4NTBkMmFiNjM0MzBhMzQ1MTgxYTliYTA5MjdkMmQ.-k4k\"><a href=\"http:\/\/avon-chemistry.com\/solution_lec_p2.html\"> http:\/\/avon-chemistry.com\/solution_lec_p2.html<\/a><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-YTAxYjMxN2QxZWQ3NmUwYTBmNGI4MDgzNGQ1OTQwYmU.-5yv\">\r\n \t<li>Why do ionic materials change the colligative properties of a solution?<\/li>\r\n \t<li>Would HCl be expected to alter colligative properties?<\/li>\r\n \t<li>Calcium carbonate is ionic, but insoluble in water. What effect would it have on the boiling point of water?<\/li>\r\n<\/ol>\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-ODk0MjAzMDJjZDlmZTU2MDg0NjMyZDU0OWRjOTRiODk.-ief\">\n<li>Perform calculations involving the effects of electrolytes on colligative properties.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>What effect do ions have?<\/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\/19212505\/20140811155609593936.jpeg\" alt=\"The presence of ions in solutions affect the properties of that solution\" width=\"400\" height=\"375\" \/><\/p>\n<p class=\"wp-caption-text\">Ionic solution measurement. Image from <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Ionic_solution_measure.jpg\">Wikimedia<\/a>.<\/p>\n<\/div>\n<p id=\"x-ck12-MjNjZjZiOTg2ZjViNzFhMzc1YWRmZTJiYTgxOTM3OGQ.-ghp\">The addition of ions creates significant changes in properties of solutions. Water molecules surround the ions and are somewhat tightly bound to them. Colligative properties are affected because the solvent properties are no longer the same as those in the pure solvent.<\/p>\n<\/div>\n<h2>Electrolytes and Colligative Properties<\/h2>\n<p id=\"x-ck12-ZGFlMjM2M2I4NjQwOWU5MGZkZjZjMDQ3YzgyNDEzNWU.-brr\">Ionic compounds are electrolytes and dissociate into two or more ions as they dissolve. This must be taken into account when calculating the freezing and boiling points of electrolyte solutions. The sample problem below demonstrates how to calculate the freezing point and boiling point of a solution of calcium chloride. Calcium chloride dissociates into three ions according to the equation:<\/p>\n<p id=\"x-ck12-ypv\"><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\/19212507\/85c2e5b28a3943bea0f1cd919f987b18.png\" alt=\"text{CaCl}_2 (s) rightarrow text{Ca}^{2+} (aq)+2text{Cl}^-(aq)\" width=\"259\" height=\"21\" \/><\/p>\n<p id=\"x-ck12-ZWRjNWVkMmFiODdmNzEyZjYwYjY4YTY0OTE5MjlmMDE.-yse\">The values of the freezing point depression and the boiling point elevation for a solution of CaCl <sub> 2 <\/sub> will be three times greater than they would be for an equal molality of a nonelectrolyte.<\/p>\n<h3>Sample Problem: Freezing and Boiling Point of an Electrolyte<\/h3>\n<p id=\"x-ck12-MDlmZDZiOThjYTZkZDIwNWFkZjU3MWFmOTk2ZTFlMTY.-vad\">Determine the freezing and boiling point of a solution prepared by dissolving 82.20 g of calcium chloride into 400.g of water.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-shk\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-h5d\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-NjJiMDRkYWE5MTgzNTQzNTdiMTlkOGIyMjY3ZTY4ZTQ.-so3\">\n<li>mass CaCl <sub> 2 <\/sub> = 82.20 g<\/li>\n<li>molar mass CaCl <sub> 2 <\/sub> = 110.98 g\/mol<\/li>\n<li>mass H <sub> 2 <\/sub> O = 400. g = 0.400 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\/19212508\/122575991dc39025b18d03758f3d1ce4.png\" alt=\"K_f(text{H}_2text{O})=-1.86^circ text{C}\/ m\" width=\"193\" height=\"20\" \/><\/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\/19212509\/58216730de76ccbd7e9e882e3169b515.png\" alt=\"K_b(text{H}_2text{O})=0.512^circ text{C}\/ m\" width=\"186\" height=\"19\" \/><\/li>\n<li>CaCl <sub> 2 <\/sub> dissociates into 3 ions<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-ljk\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-0m5\">\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\/19212510\/14603dff3c635ec418dafe7bd0063a8f.png\" alt=\"T_f=? ^circ text{C}\" width=\"72\" height=\"18\" \/><\/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\/19212511\/a6045c6e411bb44f334cd9e7d13eae41.png\" alt=\"T_b=? ^circ text{C}\" width=\"70\" height=\"15\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-MDY1YjdiOTI3NTFkZWE3MjI4YWVjZGMxMjc3NDE3ZTc.-g0u\">The moles of CaCl <sub> 2 <\/sub> is first calculated, followed by the molality of the solution. The freezing and boiling points are then determined, including multiplying by 3 for the three ions.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-z3o\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-yzq\"><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\/19212511\/994717590bfed24072df301446b06071.png\" alt=\"82.20 text{ g CaCl}_2 times frac{1 text{ mol CaCl}_2}{110.98 text{ g CaCl}_2} &amp;= 0.7407 text{ mol CaCl}_2\\frac{0.7407 text{ mol CaCl}_2}{0.400 text{ kg H}_2text{O}} &amp;= 1.85 m text{CaCl}_2\" width=\"413\" height=\"90\" \/><\/p>\n<p id=\"x-ck12-mzv\"><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\/19212514\/b2c96bb7c732a35c895343a1b5695f32.png\" alt=\"Delta T_f &amp;= K_f times m times 3=-1.86^circ text{C}\/m times 1.85 m times 3=-10.3^circ text{C} &amp;&amp; T_f=-10.3^circ text{C}\\Delta T_b &amp;= K_b times m times 3=0.512^circ text{C}\/m times 1.85 m times 3=2.84^circ text{C} &amp;&amp; T_b=102.84^circ text{C}\" width=\"587\" height=\"45\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-nfv\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-Y2M4Mzg0MzYwY2FkY2RjYTVkMGQyZWEzYjVkNWVjMWM.-a9s\">Since the normal boiling point of water is 100.00\u00b0C, the calculated result for <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\/19212501\/0e43da3d9338e5a4c8e4daa908121228.png\" alt=\"Delta T_b\" width=\"31\" height=\"15\" \/> \u00a0must be added to 100.00 to find the new boiling point.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-OWYzOTU0NWU0ZTE2ODNmNWE5ZjY0OTllNjU2NGI3ODg.-wcm\">\n<li>The effect of ionization on colligative properties is described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-Y2ZkODlkZGNmOGY3MDBkMzBmYmEzMTdlYTZiODRmNGQ.-ide\">Do the practice and homework problems dealing with ionic solutions toward the end of the section on the link below:<\/p>\n<p id=\"x-ck12-YTA4NTBkMmFiNjM0MzBhMzQ1MTgxYTliYTA5MjdkMmQ.-k4k\"><a href=\"http:\/\/avon-chemistry.com\/solution_lec_p2.html\"> http:\/\/avon-chemistry.com\/solution_lec_p2.html<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-YTAxYjMxN2QxZWQ3NmUwYTBmNGI4MDgzNGQ1OTQwYmU.-5yv\">\n<li>Why do ionic materials change the colligative properties of a solution?<\/li>\n<li>Would HCl be expected to alter colligative properties?<\/li>\n<li>Calcium carbonate is ionic, but insoluble in water. What effect would it have on the boiling point of water?<\/li>\n<\/ol>\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-2922\">\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":15,"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-2922","chapter","type-chapter","status-publish","hentry"],"part":2337,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2922","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\/2922\/revisions"}],"predecessor-version":[{"id":3422,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2922\/revisions\/3422"}],"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\/2922\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2922"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2922"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2922"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2922"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}