{"id":215197,"date":"2025-05-16T12:53:29","date_gmt":"2025-05-16T12:53:29","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=215197"},"modified":"2025-05-17T02:21:17","modified_gmt":"2025-05-17T02:21:17","slug":"the-freezing-point-depression-constant-of-chloroform-is-4-70-kg-k-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/16\/the-freezing-point-depression-constant-of-chloroform-is-4-70-kg-k-mol\/","title":{"rendered":"The freezing-point depression constant of chloroform is 4.70 kg K\/mol"},"content":{"rendered":"\n<p>The freezing-point depression constant of chloroform is 4.70 kg K\/mole. When 10.52 g of a non-volatile solute are dissolved in 150.0 g of chloroform, the solution freezes exactly 2.39 \u00b0C below the normal freezing-point of chloroform. What is the molecular weight of the solute?.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To find the molecular weight (or molar mass) of the solute, we can use the formula for <strong>freezing point depression<\/strong>:<\/p>\n\n\n\n<p>$$<br>\\Delta T_f = K_f \\cdot m<br>$$<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$\\Delta T_f$ = freezing point depression<\/li>\n\n\n\n<li>$K_f$ = freezing-point depression constant of the solvent<\/li>\n\n\n\n<li>$m$ = molality of the solution (moles of solute per kg of solvent)<\/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: Gather the given data<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$K_f = 4.70 \\ \\text{kg\u00b7K\/mol}$<\/li>\n\n\n\n<li>$\\Delta T_f = 2.39^\\circ C$ (freezing point depression)<\/li>\n\n\n\n<li>Mass of solute = 10.52 g<\/li>\n\n\n\n<li>Mass of chloroform (solvent) = 150.0 g = 0.1500 kg<\/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 2: Use formula to find molality (m)<\/h3>\n\n\n\n<p>$$<br>m = \\frac{\\Delta T_f}{K_f} = \\frac{2.39}{4.70} \\approx 0.5085 \\ \\text{mol\/kg}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Use molality to find moles of solute<\/h3>\n\n\n\n<p>Since molality = moles of solute \/ kg of solvent:<\/p>\n\n\n\n<p>$$<br>\\text{moles of solute} = m \\times \\text{kg of solvent} = 0.5085 \\times 0.1500 \\approx 0.07628 \\ \\text{mol}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate molecular weight<\/h3>\n\n\n\n<p>Molecular weight (M) is:<\/p>\n\n\n\n<p>$$<br>M = \\frac{\\text{mass of solute (g)}}{\\text{moles of solute (mol)}} = \\frac{10.52}{0.07628} \\approx 137.9 \\ \\text{g\/mol}<br>$$<\/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>$$<br>\\boxed{137.9 \\ \\text{g\/mol}}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300+ words):<\/h3>\n\n\n\n<p>This problem involves <strong>colligative properties<\/strong>, which depend on the number of solute particles in a solvent rather than the nature of the solute itself. One such property is <strong>freezing point depression<\/strong>, where the addition of a solute lowers the freezing point of the solvent.<\/p>\n\n\n\n<p>The relationship is expressed through the formula:<\/p>\n\n\n\n<p>$$<br>\\Delta T_f = K_f \\cdot m<br>$$<\/p>\n\n\n\n<p>Where $\\Delta T_f$ is the change in freezing point, $K_f$ is the freezing point depression constant for the solvent, and $m$ is the molality of the solution. In this scenario, the freezing point of chloroform decreases by 2.39\u00b0C, and the depression constant $K_f$ is 4.70 kg\u00b7K\/mol.<\/p>\n\n\n\n<p>By dividing the change in temperature by the constant, we find the molality to be approximately 0.5085 mol\/kg. Molality measures the number of moles of solute per kilogram of solvent, and with 150.0 g (or 0.1500 kg) of chloroform, we calculate the number of moles of solute to be approximately 0.07628 mol.<\/p>\n\n\n\n<p>Next, using the known mass of the solute (10.52 g), we determine its molecular weight by dividing this mass by the number of moles. This gives a final value of <strong>137.9 g\/mol<\/strong>.<\/p>\n\n\n\n<p>This technique is widely used in chemistry to identify unknown substances or to confirm the molecular formula of a compound based on its behavior in solution. It assumes ideal behavior and that the solute is non-volatile and does not dissociate.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The freezing-point depression constant of chloroform is 4.70 kg K\/mole. When 10.52 g of a non-volatile solute are dissolved in 150.0 g of chloroform, the solution freezes exactly 2.39 \u00b0C below the normal freezing-point of chloroform. What is the molecular weight of the solute?. The Correct Answer and Explanation is: To find the molecular weight [&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-215197","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215197","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=215197"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215197\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=215197"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=215197"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=215197"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}