{"id":213992,"date":"2025-05-13T09:08:53","date_gmt":"2025-05-13T09:08:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213992"},"modified":"2025-05-13T09:08:55","modified_gmt":"2025-05-13T09:08:55","slug":"the-freezing-point-tf-for-t-butanol-is-25-508c-and-kf-is-9-18c-m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/13\/the-freezing-point-tf-for-t-butanol-is-25-508c-and-kf-is-9-18c-m\/","title":{"rendered":"The freezing point (Tf) for t-butanol is 25.508C and Kf is 9.18C\/m."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The freezing point (Tf) for t-butanol is 25.508C and Kf is 9.18C\/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 10.0-g sample t-butanol is measured as 24.598C, how many grams of water are present in the sample?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[A] 10 g<br>[B] 1.8 g<br>[C] 0.10 g<br>[D] 0.018 g<br>[E] 18. g<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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 class=\"wp-block-paragraph\">To solve this problem, we use <strong>freezing point depression<\/strong>, a <strong>colligative property<\/strong> described by the equation: \u0394Tf=Kf\u22c5m\\Delta T_f = K_f \\cdot m<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">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>KfK_f = freezing point depression constant (\u00b0C\/m)<\/li>\n\n\n\n<li>mm = molality of solute (mol\/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\"><strong>Step 1: Determine \u0394Tf\\Delta T_f<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">\u0394Tf=Tf,pure\u2212Tf,solution=25.508\u2218C\u221224.598\u2218C=0.910\u2218C\\Delta T_f = T_{f,\\text{pure}} &#8211; T_{f,\\text{solution}} = 25.508^\\circ C &#8211; 24.598^\\circ C = 0.910^\\circ C<\/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 2: Solve for molality (m)<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">m=\u0394TfKf=0.9109.18=0.09913&nbsp;mol\/kgm = \\frac{\\Delta T_f}{K_f} = \\frac{0.910}{9.18} = 0.09913\\ \\text{mol\/kg}<\/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 3: Calculate moles of water in the sample<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We are told that the total mass of <strong>t-butanol<\/strong> is 10.0 g. Water is the <strong>impurity (solute)<\/strong>, and t-butanol is the <strong>solvent<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Molality mm is defined as: m=moles&nbsp;of&nbsp;solute&nbsp;(water)kg&nbsp;of&nbsp;solvent&nbsp;(t-butanol)m = \\frac{\\text{moles of solute (water)}}{\\text{kg of solvent (t-butanol)}}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So: moles&nbsp;of&nbsp;water=m\u00d7kg&nbsp;of&nbsp;t-butanol=0.09913\u00d710.01000=9.913\u00d710\u22124&nbsp;mol\\text{moles of water} = m \\times \\text{kg of t-butanol} = 0.09913 \\times \\frac{10.0}{1000} = 9.913 \\times 10^{-4}\\ \\text{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>Step 4: Convert moles of water to grams<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Molar mass of water = 18.015 g\/mol mass&nbsp;of&nbsp;water=9.913\u00d710\u22124\u00d718.015=0.01784&nbsp;g\\text{mass of water} = 9.913 \\times 10^{-4} \\times 18.015 = 0.01784\\ \\text{g}<\/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 class=\"wp-block-paragraph\">0.018&nbsp;g\u21d2Choice&nbsp;[D]\\boxed{0.018\\ \\text{g}} \\Rightarrow \\text{Choice [D]}<\/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 (Summary)<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This problem deals with <strong>freezing point depression<\/strong>, where the presence of solute particles (in this case, <strong>water<\/strong>) lowers the freezing point of a solvent (here, <strong>t-butanol<\/strong>). When exposed to air, t-butanol absorbs water, acting as a solute that disrupts the freezing process. The key to solving the problem is calculating how much water is needed to lower the freezing point by 0.910\u00b0C using the known constant Kf=9.18\u00b0C\/mK_f = 9.18\u00b0C\/m. After finding the molality, it\u2019s used with the mass of the t-butanol to determine how many moles (and then grams) of water were present. The precise calculation yields approximately <strong>0.018 g<\/strong>, confirming that <strong>Choice [D]<\/strong> is correct.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The freezing point (Tf) for t-butanol is 25.508C and Kf is 9.18C\/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 10.0-g sample t-butanol is measured as 24.598C, how many grams of water are present in the sample? [A] 10 g[B] 1.8 g[C] 0.10 g[D] 0.018 g[E] 18. g [&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-213992","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213992","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=213992"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213992\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213992"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213992"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213992"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}