{"id":182239,"date":"2025-01-13T14:18:37","date_gmt":"2025-01-13T14:18:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182239"},"modified":"2025-01-13T14:18:40","modified_gmt":"2025-01-13T14:18:40","slug":"biphenyl-c12h10-molar-mass-154-22-g-mol-is-a-non-volatile-nonionizing-solute-that-is-soluble-in-benzene","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/biphenyl-c12h10-molar-mass-154-22-g-mol-is-a-non-volatile-nonionizing-solute-that-is-soluble-in-benzene\/","title":{"rendered":"Biphenyl, C12?H10 (molar mass =154.22 g\/mol ), is a non-volatile, nonionizing solute that is soluble in benzene"},"content":{"rendered":"\n<p>Biphenyl, C12?H10 (molar mass =154.22 g\/mol ), is a non-volatile, nonionizing solute that is soluble in benzene, C6?H6? (molar mass =78,12 g\/mol ). At 25\u00b0C, the vapour pressure of pure benzene is 100.84 tort. What is the vapour pressure, in tort, of a solution made from dissolving 15.2 g of biphenyl in 31.3 g of benzene? Vapour pressure = tort. Report your answer to three significant figures. Do NOT include units in your answer.<\/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 find the vapor pressure of the solution, we can use <strong>Raoult&#8217;s Law<\/strong>:<\/p>\n\n\n\n<p>[<br>P_{\\text{solution}} = X_{\\text{solvent}} \\cdot P_{\\text{solvent}}^{\\text{pure}}<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( P_{\\text{solution}} ) is the vapor pressure of the solution.<\/li>\n\n\n\n<li>( X_{\\text{solvent}} ) is the mole fraction of the solvent (benzene) in the solution.<\/li>\n\n\n\n<li>( P_{\\text{solvent}}^{\\text{pure}} ) is the vapor pressure of the pure solvent (benzene).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the moles of biphenyl (solute) and benzene (solvent).<\/h3>\n\n\n\n<p><strong>Moles of biphenyl:<\/strong><\/p>\n\n\n\n<p>[<br>\\text{moles of biphenyl} = \\frac{\\text{mass of biphenyl}}{\\text{molar mass of biphenyl}} = \\frac{15.2\\ \\text{g}}{154.22\\ \\text{g\/mol}} = 0.0985\\ \\text{mol}<br>]<\/p>\n\n\n\n<p><strong>Moles of benzene:<\/strong><\/p>\n\n\n\n<p>[<br>\\text{moles of benzene} = \\frac{\\text{mass of benzene}}{\\text{molar mass of benzene}} = \\frac{31.3\\ \\text{g}}{78.12\\ \\text{g\/mol}} = 0.400\\ \\text{mol}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the mole fraction of benzene.<\/h3>\n\n\n\n<p>The mole fraction of benzene ((X_{\\text{benzene}})) is:<\/p>\n\n\n\n<p>[<br>X_{\\text{benzene}} = \\frac{\\text{moles of benzene}}{\\text{moles of benzene} + \\text{moles of biphenyl}} = \\frac{0.400}{0.400 + 0.0985} = \\frac{0.400}{0.4985} = 0.802<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Apply Raoult&#8217;s Law.<\/h3>\n\n\n\n<p>Now, use Raoult&#8217;s Law to find the vapor pressure of the solution:<\/p>\n\n\n\n<p>[<br>P_{\\text{solution}} = X_{\\text{benzene}} \\cdot P_{\\text{benzene}}^{\\text{pure}} = 0.802 \\cdot 100.84\\ \\text{torr} = 80.9\\ \\text{torr}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The vapor pressure of the solution is <strong>80.9 torr<\/strong> (to three significant figures).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>This calculation assumes ideal behavior of the solution (no significant deviations from Raoult&#8217;s Law). The mole fraction of benzene, which is the solvent, influences the vapor pressure. As biphenyl is non-volatile, it does not contribute to the vapor pressure, so the solution&#8217;s vapor pressure depends solely on the benzene. By using Raoult&#8217;s Law, we adjusted the vapor pressure of pure benzene based on its mole fraction in the solution.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Biphenyl, C12?H10 (molar mass =154.22 g\/mol ), is a non-volatile, nonionizing solute that is soluble in benzene, C6?H6? (molar mass =78,12 g\/mol ). At 25\u00b0C, the vapour pressure of pure benzene is 100.84 tort. What is the vapour pressure, in tort, of a solution made from dissolving 15.2 g of biphenyl in 31.3 g 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-182239","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182239","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=182239"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182239\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182239"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182239"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}