{"id":222301,"date":"2025-05-31T08:17:54","date_gmt":"2025-05-31T08:17:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222301"},"modified":"2025-05-31T08:17:56","modified_gmt":"2025-05-31T08:17:56","slug":"when-48-7g-of-benzamide-c7h7no-are-dissolved-in-1300-g-of-a-certain-mystery-liquid-x-the-freezing-point-of-the-solution-is-2-3aac-lower-than-the-freezing-point-of-pure-x","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/when-48-7g-of-benzamide-c7h7no-are-dissolved-in-1300-g-of-a-certain-mystery-liquid-x-the-freezing-point-of-the-solution-is-2-3aac-lower-than-the-freezing-point-of-pure-x\/","title":{"rendered":"When 48.7g of benzamide C7H7NO are dissolved in 1300.g of a certain mystery liquid X, the freezing point of the solution is 2.3\u00c3\u201a\u00c2\u00b0C lower than the freezing point of pure X."},"content":{"rendered":"\n<p>When 48.7g of benzamide C7H7NO are dissolved in 1300.g of a certain mystery liquid X, the freezing point of the solution is 2.3\u00c3\u201a\u00c2\u00b0C lower than the freezing point of pure X. On the other hand, when 48.7g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 8.7\u00c3\u201a\u00c2\u00b0C lower than the freezing point of pure X.Calculate the van&#8217;t Hoff factor for sodium chloride inX. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits. Please dont round untill the end.<\/p>\n\n\n\n<p>i = \u00c3\u201a\u00c2<br>\u00c3\u201a\u00c2<\/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 (i)<\/strong> for <strong>sodium chloride (NaCl)<\/strong> in solvent X, we compare the <strong>freezing point depression<\/strong> it causes with that caused by <strong>benzamide<\/strong>, a <strong>non-electrolyte<\/strong> (i = 1).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-Step Calculation<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>1. Freezing Point Depression Equation<\/strong>:<\/h4>\n\n\n\n<p>\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 = freezing point depression (\u00b0C)<\/li>\n\n\n\n<li>ii = van\u2019t Hoff factor<\/li>\n\n\n\n<li>KfK_f = cryoscopic constant (same for both solutions)<\/li>\n\n\n\n<li>mm = molality of solute<\/li>\n<\/ul>\n\n\n\n<p>We will use benzamide to determine KfK_f because its van\u2019t Hoff factor is known (i = 1).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>For Benzamide:<\/strong><\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of benzamide = 48.7 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 = 1300 g = 1.300 kg<\/li>\n\n\n\n<li>\u0394Tf=2.3\u2218C\\Delta T_f = 2.3^\\circ C<\/li>\n<\/ul>\n\n\n\n<p><strong>Molality (m):<\/strong> mol&nbsp;benzamide=48.7121.14\u22480.402mol\\text{mol benzamide} = \\frac{48.7}{121.14} \\approx 0.402 mol m=0.4021.300\u22480.309\u2009mol\/kgm = \\frac{0.402}{1.300} \\approx 0.309 \\, \\text{mol\/kg}<\/p>\n\n\n\n<p>Now use: \u0394Tf=i\u22c5Kf\u22c5m=1\u22c5Kf\u22c50.309=2.3\\Delta T_f = i \\cdot K_f \\cdot m = 1 \\cdot K_f \\cdot 0.309 = 2.3 K_f = \\frac{2.3}{0.309} \\approx 7.44 \\, ^\\circ C \\cdot \\text{kg\/mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>For NaCl:<\/strong><\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass NaCl = 48.7 g<\/li>\n\n\n\n<li>Molar mass NaCl \u2248 58.44 g\/mol<\/li>\n\n\n\n<li>\u0394Tf=8.7\u2218C\\Delta T_f = 8.7^\\circ C<\/li>\n<\/ul>\n\n\n\n<p><strong>Molality:<\/strong> mol&nbsp;NaCl=48.758.44\u22480.833mol\\text{mol NaCl} = \\frac{48.7}{58.44} \\approx 0.833 mol m=0.8331.300\u22480.641\u2009mol\/kgm = \\frac{0.833}{1.300} \\approx 0.641 \\, \\text{mol\/kg}<\/p>\n\n\n\n<p>Now solve for ii: \u0394Tf=i\u22c5Kf\u22c5m\\Delta T_f = i \\cdot K_f \\cdot m 8.7=i\u22c57.44\u22c50.6418.7 = i \\cdot 7.44 \\cdot 0.641 i=8.77.44\u22c50.641\u22488.74.77\u22481.823i = \\frac{8.7}{7.44 \\cdot 0.641} \\approx \\frac{8.7}{4.77} \\approx 1.823<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer<\/strong>:<\/h3>\n\n\n\n<p>i=1.8\\boxed{i = 1.8}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The <strong>van\u2019t Hoff factor (i)<\/strong> quantifies how many particles a solute produces in solution, and is essential for calculating colligative properties like <strong>freezing point depression<\/strong>. For <strong>non-electrolytes<\/strong> like benzamide, i = 1 because they do <strong>not dissociate<\/strong> in solution. <strong>Ionic compounds<\/strong>, such as <strong>NaCl<\/strong>, dissociate into ions\u2014in this case, ideally into Na\u207a and Cl\u207b, suggesting a theoretical i = 2. However, in real solutions, <strong>ion pairing<\/strong> and <strong>incomplete dissociation<\/strong> reduce the effective van\u2019t Hoff factor.<\/p>\n\n\n\n<p>In this problem, both benzamide and NaCl are dissolved in the same mass of solvent X. We first calculate the <strong>molality<\/strong> of each solute. Using the known freezing point depression for benzamide and the assumption that it behaves ideally (i = 1), we determine the <strong>freezing point depression constant<\/strong> KfK_f for the solvent.<\/p>\n\n\n\n<p>Then, using the same KfK_f, we calculate the molality of NaCl and plug these into the freezing point depression formula to solve for the van\u2019t Hoff factor. The result, i\u22481.8i \\approx 1.8, shows that NaCl <strong>mostly but not completely dissociates<\/strong>, which is typical in non-ideal solutions. This effective van\u2019t Hoff factor reflects real-world conditions like <strong>ion-ion interactions<\/strong> or <strong>solvent effects<\/strong> that reduce dissociation.<\/p>\n\n\n\n<p>This approach allows us to experimentally determine how many particles are influencing the colligative property, providing insight into the <strong>ionic behavior<\/strong> of solutes in solution.<\/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-banner4-188.jpeg\" alt=\"\" class=\"wp-image-222302\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>When 48.7g of benzamide C7H7NO are dissolved in 1300.g of a certain mystery liquid X, the freezing point of the solution is 2.3\u00c3\u201a\u00c2\u00b0C lower than the freezing point of pure X. On the other hand, when 48.7g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is [&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-222301","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222301","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=222301"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222301\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222301"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222301"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222301"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}