{"id":233643,"date":"2025-06-13T09:30:31","date_gmt":"2025-06-13T09:30:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233643"},"modified":"2025-06-13T09:30:34","modified_gmt":"2025-06-13T09:30:34","slug":"biphenyl-c12h10-is-a-nonvolatile-nonionizing-solute-that-is-soluble-in-benzene-c6h6","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/biphenyl-c12h10-is-a-nonvolatile-nonionizing-solute-that-is-soluble-in-benzene-c6h6\/","title":{"rendered":"Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6."},"content":{"rendered":"\n<p>Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 \u00c2\u00b0C, the vapor pressure of pure benzene is 100.84 Torr. What is the vapor pressure of a solution made from dissolving 13.7 g of biphenyl in 28.1 g of benzene? Psolution = <em>__<\/em> Torr<\/p>\n\n\n\n<p><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>To calculate the vapor pressure of a solution of biphenyl in benzene, we use <strong>Raoult\u2019s Law<\/strong>, which is: Psolution=Xsolvent\u22c5Psolvent0P_{\\text{solution}} = X_{\\text{solvent}} \\cdot P^0_{\\text{solvent}}Psolution\u200b=Xsolvent\u200b\u22c5Psolvent0\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>PsolutionP_{\\text{solution}}Psolution\u200b is the vapor pressure of the solution.<\/li>\n\n\n\n<li>XsolventX_{\\text{solvent}}Xsolvent\u200b is the mole fraction of the solvent (benzene).<\/li>\n\n\n\n<li>Psolvent0P^0_{\\text{solvent}}Psolvent0\u200b is the vapor pressure of the pure solvent (benzene), which is given as 100.84 Torr.<\/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: Molar masses<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Biphenyl (C\u2081\u2082H\u2081\u2080)<\/strong>:<br>C = 12.01, H = 1.008<br>Molar mass = (12 \u00d7 12.01) + (10 \u00d7 1.008) = 144.12 + 10.08 = <strong>154.20 g\/mol<\/strong><\/li>\n\n\n\n<li><strong>Benzene (C\u2086H\u2086)<\/strong>:<br>Molar mass = (6 \u00d7 12.01) + (6 \u00d7 1.008) = 72.06 + 6.048 = <strong>78.11 g\/mol<\/strong><\/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 2: Moles of solute and solvent<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of biphenyl = 13.7 g \u00f7 154.20 g\/mol \u2248 <strong>0.0889 mol<\/strong><\/li>\n\n\n\n<li>Moles of benzene = 28.1 g \u00f7 78.11 g\/mol \u2248 <strong>0.3596 mol<\/strong><\/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 3: Mole fraction of benzene<\/strong><\/h3>\n\n\n\n<p>Xbenzene=nbenzenenbenzene+nbiphenyl=0.35960.3596+0.0889\u22480.35960.4485\u22480.8018X_{\\text{benzene}} = \\frac{n_{\\text{benzene}}}{n_{\\text{benzene}} + n_{\\text{biphenyl}}} = \\frac{0.3596}{0.3596 + 0.0889} \u2248 \\frac{0.3596}{0.4485} \u2248 0.8018Xbenzene\u200b=nbenzene\u200b+nbiphenyl\u200bnbenzene\u200b\u200b=0.3596+0.08890.3596\u200b\u22480.44850.3596\u200b\u22480.8018<\/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: Apply Raoult\u2019s Law<\/strong><\/h3>\n\n\n\n<p>Psolution=Xbenzene\u22c5Pbenzene0=0.8018\u22c5100.84\u224880.91&nbsp;TorrP_{\\text{solution}} = X_{\\text{benzene}} \\cdot P^0_{\\text{benzene}} = 0.8018 \\cdot 100.84 \u2248 80.91 \\text{ Torr}Psolution\u200b=Xbenzene\u200b\u22c5Pbenzene0\u200b=0.8018\u22c5100.84\u224880.91&nbsp;Torr<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>80.91&nbsp;Torr\\boxed{80.91\\ \\text{Torr}}80.91&nbsp;Torr\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udcd8 <strong>Explanation (300 words)<\/strong><\/h3>\n\n\n\n<p>Raoult\u2019s Law governs how a nonvolatile solute affects the vapor pressure of a solvent. Since biphenyl does not vaporize (nonvolatile) and does not ionize in solution (nonionizing), it only <em>lowers<\/em> the vapor pressure of benzene by reducing the number of benzene molecules available at the surface to evaporate.<\/p>\n\n\n\n<p>The vapor pressure of a liquid is due to molecules at the surface escaping into the vapor phase. When a solute is dissolved in the solvent, some surface area is occupied by the solute particles, which do not evaporate. As a result, fewer solvent molecules can escape into the gas phase, and the overall vapor pressure decreases.<\/p>\n\n\n\n<p>To quantify this, we calculate the <strong>mole fraction of the solvent<\/strong>, which is the ratio of solvent moles to total moles in solution. For this problem, we determined the moles of benzene and biphenyl using their molar masses and then used this to find the mole fraction of benzene.<\/p>\n\n\n\n<p>Multiplying this mole fraction by the vapor pressure of pure benzene gives the new, lower vapor pressure of the solution. This reflects how the presence of biphenyl lowers benzene&#8217;s vapor pressure proportionally to how much biphenyl is added.<\/p>\n\n\n\n<p>This is an ideal solution approximation, assuming that the interactions between benzene and biphenyl molecules are similar to those between pure benzene molecules. This approach is common in introductory chemistry to demonstrate colligative properties, which depend only on the number of solute particles, 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\/06\/learnexams-banner8-292.jpeg\" alt=\"\" class=\"wp-image-233644\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 \u00c2\u00b0C, the vapor pressure of pure benzene is 100.84 Torr. What is the vapor pressure of a solution made from dissolving 13.7 g of biphenyl in 28.1 g of benzene? Psolution = __ Torr The Correct Answer and Explanation is: [&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-233643","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233643","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=233643"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233643\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233643"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233643"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233643"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}