{"id":222305,"date":"2025-05-31T08:20:52","date_gmt":"2025-05-31T08:20:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222305"},"modified":"2025-05-31T08:20:54","modified_gmt":"2025-05-31T08:20:54","slug":"when-151-g-of-benzamide-c7h7no-are-dissolved-in-1200-g-of-a-certain-mystery-liquid-x-the-freezing-point-of-the-solution-is-8-0-c-lower-than-the-freezing-point-of-pure-x","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/when-151-g-of-benzamide-c7h7no-are-dissolved-in-1200-g-of-a-certain-mystery-liquid-x-the-freezing-point-of-the-solution-is-8-0-c-lower-than-the-freezing-point-of-pure-x\/","title":{"rendered":"When 151. g of benzamide C7H7NO are dissolved in 1200. g of a certain mystery liquid X, the freezing point of the solution is 8.0 \u00b0C lower than the freezing point of pure X"},"content":{"rendered":"\n<p>When 151. g of benzamide C7H7NO are dissolved in 1200. g of a certain mystery liquid X, the freezing point of the solution is 8.0 \u00b0C lower than the freezing point of pure X. On the other hand, when 151. g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 20.5\u00b0C lower than the freezing point of pure X. Calculate the van&#8217;t Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the <strong>van\u2019t Hoff factor<\/strong> (<em>i<\/em>) for <strong>iron(III) chloride<\/strong> (FeCl\u2083) in solvent X, we&#8217;ll use <strong>freezing point depression<\/strong>: \u0394Tf=i\u22c5Kf\u22c5m\\Delta T_f = i \\cdot K_f \\cdot m<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Tf\\Delta T_f is the freezing point depression,<\/li>\n\n\n\n<li>ii is the van\u2019t Hoff factor,<\/li>\n\n\n\n<li>KfK_f is the cryoscopic constant (same for both solutions since the solvent is the same),<\/li>\n\n\n\n<li>mm is the molality of the solution.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Use benzamide to determine KfK_f<\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of benzamide = 151 g<\/li>\n\n\n\n<li>Molar mass of benzamide (C\u2087H\u2087NO) \u2248 121.14 g\/mol<\/li>\n\n\n\n<li>Mass of solvent X = 1200. g = 1.200 kg<\/li>\n\n\n\n<li>Freezing point depression = 8.0\u00b0C<\/li>\n\n\n\n<li>Benzamide is a <strong>non-electrolyte<\/strong>, so i=1i = 1<\/li>\n<\/ul>\n\n\n\n<p><strong>Molality:<\/strong> m=151121.14\u00f71.200=1.039&nbsp;mol\/kgm = \\frac{151}{121.14} \\div 1.200 = 1.039 \\text{ mol\/kg}<\/p>\n\n\n\n<p>Using the formula: 8.0=1\u22c5Kf\u22c51.039\u21d2Kf=8.01.039\u22487.7&nbsp;\u00b0C\\cdotpkg\/mol8.0 = 1 \\cdot K_f \\cdot 1.039 \\Rightarrow K_f = \\frac{8.0}{1.039} \u2248 7.7 \\text{ \u00b0C\u00b7kg\/mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use FeCl\u2083 data to find ii<\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of FeCl\u2083 = 151 g<\/li>\n\n\n\n<li>Molar mass of FeCl\u2083 \u2248 162.20 g\/mol<\/li>\n\n\n\n<li>Mass of solvent X = 1.200 kg<\/li>\n\n\n\n<li>Freezing point depression = 20.5\u00b0C<\/li>\n<\/ul>\n\n\n\n<p><strong>Molality:<\/strong> m=151162.20\u00f71.200=0.7749&nbsp;mol\/kgm = \\frac{151}{162.20} \\div 1.200 = 0.7749 \\text{ mol\/kg}<\/p>\n\n\n\n<p>Now solve for ii: 20.5=i\u22c57.7\u22c50.7749\u21d2i=20.57.7\u22c50.7749\u224820.55.9667\u22483.420.5 = i \\cdot 7.7 \\cdot 0.7749 \\Rightarrow i = \\frac{20.5}{7.7 \\cdot 0.7749} \u2248 \\frac{20.5}{5.9667} \u2248 3.4<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>i=3.4&nbsp;(unitless)\\boxed{i = 3.4\\ (\\text{unitless})}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>The van\u2019t Hoff factor (<em>i<\/em>) represents the number of particles into which a compound dissociates in solution. For <strong>non-electrolytes<\/strong> like benzamide, which don\u2019t dissociate, <em>i<\/em> = 1. For <strong>ionic compounds<\/strong> like <strong>FeCl\u2083<\/strong>, which dissociate into ions, <em>i<\/em> can be more than 1. The theoretical maximum for FeCl\u2083 is 4, since it dissociates into Fe\u00b3\u207a and 3 Cl\u207b ions.<\/p>\n\n\n\n<p>To calculate <em>i<\/em> experimentally, we compare the freezing point depression (\u0394Tf\\Delta T_f) of a known non-electrolyte and the ionic compound in the same solvent. The formula \u0394Tf=i\u22c5Kf\u22c5m\\Delta T_f = i \\cdot K_f \\cdot m links the depression to molality and the van\u2019t Hoff factor. We first used benzamide\u2019s data to find the cryoscopic constant KfK_f of solvent X, since <em>i<\/em> = 1 for benzamide. Then, using the same solvent and mass, we determined the molality of the FeCl\u2083 solution and plugged everything into the formula to solve for <em>i<\/em>.<\/p>\n\n\n\n<p>We found the experimental van\u2019t Hoff factor for FeCl\u2083 to be <strong>3.4<\/strong>, slightly lower than the ideal value of 4. This suggests <strong>incomplete dissociation<\/strong> or <strong>ion-pairing<\/strong> in the solvent, common in non-aqueous or less polar solvents. The difference highlights real-world behavior versus ideal dissociation assumptions.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner5-162.jpeg\" alt=\"\" class=\"wp-image-222306\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>When 151. g of benzamide C7H7NO are dissolved in 1200. g of a certain mystery liquid X, the freezing point of the solution is 8.0 \u00b0C lower than the freezing point of pure X. On the other hand, when 151. g of iron(III) chloride are dissolved in the same mass of X, the freezing point [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-222305","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222305","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=222305"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222305\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222305"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222305"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222305"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}