{"id":181899,"date":"2025-01-11T17:24:24","date_gmt":"2025-01-11T17:24:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=181899"},"modified":"2025-01-11T17:24:26","modified_gmt":"2025-01-11T17:24:26","slug":"the-freezing-point-for-pure-lauric-acid-is-43-8-c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/11\/the-freezing-point-for-pure-lauric-acid-is-43-8-c\/","title":{"rendered":"The freezing point for pure lauric acid is 43.8 \u00b0C"},"content":{"rendered":"\n<p>The freezing point for pure lauric acid is 43.8 \u00b0C. When 8.00 g of lauric acid is mixed with 1.00 g of benzoic acid, the freezing point drops to 39.2 \u00b0C.<\/p>\n\n\n\n<p>What is the experimental value for the molality of the mixture if Kfp for lauric acid is \u20133.9 \u00b0C\/m?<br>What is the number of moles of the solute?<br>What is the experimental molar mass of the solute?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To solve this, we need to calculate the experimental molality, moles of the solute, and the experimental molar mass of the solute.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the change in freezing point (\u0394Tf)<\/h3>\n\n\n\n<p>The freezing point depression formula is:<\/p>\n\n\n\n<p>[<br>\\Delta T_f = K_f \\times m<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(\\Delta T_f) is the freezing point depression,<\/li>\n\n\n\n<li>(K_f) is the freezing point depression constant of the solvent (lauric acid),<\/li>\n\n\n\n<li>(m) is the molality of the solution.<\/li>\n<\/ul>\n\n\n\n<p>First, calculate the change in freezing point:<\/p>\n\n\n\n<p>[<br>\\Delta T_f = T_{\\text{freeze, solvent}} &#8211; T_{\\text{freeze, solution}}<br>]<br>[<br>\\Delta T_f = 43.8^\\circ C &#8211; 39.2^\\circ C = 4.6^\\circ C<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate molality (m)<\/h3>\n\n\n\n<p>Now that we have the freezing point depression, we can rearrange the freezing point depression formula to solve for molality (m):<\/p>\n\n\n\n<p>[<br>m = \\frac{\\Delta T_f}{K_f}<br>]<br>[<br>m = \\frac{4.6^\\circ C}{3.9^\\circ C\/m} = 1.18 \\, \\text{mol\/kg}<br>]<\/p>\n\n\n\n<p>So, the experimental molality is 1.18 mol\/kg.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the moles of solute (benzoic acid)<\/h3>\n\n\n\n<p>Next, we use the definition of molality:<\/p>\n\n\n\n<p>[<br>m = \\frac{\\text{moles of solute}}{\\text{mass of solvent (kg)}}<br>]<\/p>\n\n\n\n<p>We know that:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molality (m = 1.18 \\, \\text{mol\/kg}),<\/li>\n\n\n\n<li>Mass of the solvent (lauric acid) = 8.00 g = 0.00800 kg.<\/li>\n<\/ul>\n\n\n\n<p>Rearranging the equation to solve for moles of solute:<\/p>\n\n\n\n<p>[<br>\\text{moles of solute} = m \\times \\text{mass of solvent (kg)}<br>]<br>[<br>\\text{moles of solute} = 1.18 \\, \\text{mol\/kg} \\times 0.00800 \\, \\text{kg} = 0.00944 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p>So, the moles of benzoic acid (solute) are 0.00944 mol.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate the experimental molar mass of the solute<\/h3>\n\n\n\n<p>Finally, we can calculate the experimental molar mass of the solute using:<\/p>\n\n\n\n<p>[<br>\\text{Molar mass} = \\frac{\\text{mass of solute (g)}}{\\text{moles of solute (mol)}}<br>]<br>[<br>\\text{Molar mass} = \\frac{1.00 \\, \\text{g}}{0.00944 \\, \\text{mol}} = 105.9 \\, \\text{g\/mol}<br>]<\/p>\n\n\n\n<p>Thus, the experimental molar mass of the solute (benzoic acid) is <strong>105.9 g\/mol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Results:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molality<\/strong> = 1.18 mol\/kg<\/li>\n\n\n\n<li><strong>Moles of solute<\/strong> = 0.00944 mol<\/li>\n\n\n\n<li><strong>Experimental molar mass of the solute<\/strong> = 105.9 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>This approach uses the concept of freezing point depression and molality to find the molar mass of the solute in the mixture.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The freezing point for pure lauric acid is 43.8 \u00b0C. When 8.00 g of lauric acid is mixed with 1.00 g of benzoic acid, the freezing point drops to 39.2 \u00b0C. What is the experimental value for the molality of the mixture if Kfp for lauric acid is \u20133.9 \u00b0C\/m?What is the number of moles [&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-181899","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181899","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=181899"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181899\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181899"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181899"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181899"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}