{"id":257006,"date":"2025-07-16T15:23:35","date_gmt":"2025-07-16T15:23:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257006"},"modified":"2025-07-16T15:23:37","modified_gmt":"2025-07-16T15:23:37","slug":"consider-a-solution-containing-a-mole-of-toluene-and-a-hundred-moles-of-benzene","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/16\/consider-a-solution-containing-a-mole-of-toluene-and-a-hundred-moles-of-benzene\/","title":{"rendered":"Consider a solution containing a mole of toluene and a hundred moles of benzene."},"content":{"rendered":"\n<p>Consider a solution containing a mole of toluene and a hundred moles of benzene. Kf=5.12\u00b0C\/m Kb=2.65\u00b0C\/m Melting point= 5.5\u00b0C Liquefaction point= 80.1\u00b0C What is the freezing point of the solution? A. 4.84\u00b0C B. 4.99\u00b0C C. 5.28\u00b0C D. 5.43\u00b0C<\/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><strong>Correct Answer: A. 4.84\u00b0C<\/strong><\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>To find the <strong>freezing point<\/strong> of the solution, we use the formula for <strong>freezing point depression<\/strong>: \u0394Tf=Kf\u00d7m\\Delta T_f = K_f \\times m\u0394Tf\u200b=Kf\u200b\u00d7m<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Tf\\Delta T_f\u0394Tf\u200b is the <strong>freezing point depression<\/strong><\/li>\n\n\n\n<li>KfK_fKf\u200b is the <strong>freezing point depression constant<\/strong> (for benzene, given as 5.12\u00b0C\/m)<\/li>\n\n\n\n<li>mmm is the <strong>molality<\/strong> of the solute<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 1: Identify known values<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 mole of <strong>toluene<\/strong> (solute)<\/li>\n\n\n\n<li>100 moles of <strong>benzene<\/strong> (solvent)<\/li>\n\n\n\n<li>Freezing point of <strong>pure benzene<\/strong> is 5.5\u00b0C<\/li>\n\n\n\n<li>Kf=5.12K_f = 5.12Kf\u200b=5.12\u00b0C\/m<\/li>\n<\/ul>\n\n\n\n<p>We assume the <strong>mass of the solvent<\/strong> (benzene) is known or can be estimated. Since benzene has a molar mass of about 78 g\/mol, 100 moles would be: 100&nbsp;mol\u00d778&nbsp;g\/mol=7800&nbsp;g=7.8&nbsp;kg100 \\text{ mol} \\times 78 \\text{ g\/mol} = 7800 \\text{ g} = 7.8 \\text{ kg}100&nbsp;mol\u00d778&nbsp;g\/mol=7800&nbsp;g=7.8&nbsp;kg<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 2: Calculate molality of the solution<\/strong><\/p>\n\n\n\n<p>Molality (m) is moles of solute per kg of solvent: m=1&nbsp;mol&nbsp;toluene7.8&nbsp;kg&nbsp;benzene\u22480.1282&nbsp;mol\/kgm = \\frac{1 \\text{ mol toluene}}{7.8 \\text{ kg benzene}} \\approx 0.1282 \\text{ mol\/kg}m=7.8&nbsp;kg&nbsp;benzene1&nbsp;mol&nbsp;toluene\u200b\u22480.1282&nbsp;mol\/kg<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 3: Calculate freezing point depression<\/strong> \u0394Tf=5.12\u00d70.1282\u22480.6566\u2218C\\Delta T_f = 5.12 \\times 0.1282 \\approx 0.6566^\\circ C\u0394Tf\u200b=5.12\u00d70.1282\u22480.6566\u2218C<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 4: Calculate new freezing point<\/strong> Freezing&nbsp;point=5.5\u2218C\u22120.6566\u2218C\u22484.84\u2218C\\text{Freezing point} = 5.5^\\circ C &#8211; 0.6566^\\circ C \\approx 4.84^\\circ CFreezing&nbsp;point=5.5\u2218C\u22120.6566\u2218C\u22484.84\u2218C<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Final Answer: A. 4.84\u00b0C<\/strong><\/p>\n\n\n\n<p>This result shows how adding a small amount of solute (toluene) lowers the freezing point of a solvent (benzene) due to the <strong>colligative property<\/strong> known as <strong>freezing point depression<\/strong>, which depends on the number of solute particles in the solution, not their identity.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-200.jpeg\" alt=\"\" class=\"wp-image-257008\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Consider a solution containing a mole of toluene and a hundred moles of benzene. Kf=5.12\u00b0C\/m Kb=2.65\u00b0C\/m Melting point= 5.5\u00b0C Liquefaction point= 80.1\u00b0C What is the freezing point of the solution? A. 4.84\u00b0C B. 4.99\u00b0C C. 5.28\u00b0C D. 5.43\u00b0C The Correct Answer and Explanation is: Correct Answer: A. 4.84\u00b0C Explanation: To find the freezing point of [&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-257006","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257006","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=257006"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257006\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257006"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257006"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257006"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}