{"id":182148,"date":"2025-01-13T12:49:49","date_gmt":"2025-01-13T12:49:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182148"},"modified":"2025-01-13T12:49:51","modified_gmt":"2025-01-13T12:49:51","slug":"aa-what-mass-of-ethylene-glycol-c2h6o2-molar-mass-62-1-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/aa-what-mass-of-ethylene-glycol-c2h6o2-molar-mass-62-1-g-mol\/","title":{"rendered":"\u00c3\u201a\u00c2 What mass of ethylene glycol (C2H6O2, molar mass = 62.1 g\/mol)"},"content":{"rendered":"\n<p>\u00c3\u201a\u00c2 What mass of ethylene glycol (C2H6O2, molar mass = 62.1 g\/mol), the main component of antifreeze, must be added to 8.5 L water to produce a solution for use in a car\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2s radiator that freezes at -10.0oF (-23.3oC)? Assume the density of water is exactly 1.0 g\/mL. Kf of water = 1.86 oC\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u0153kg\/mol Molar mass\u00c3\u201a\u00c2 of ethylene glycol = 62.1 g\/mole<\/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 prepare an antifreeze solution that freezes at -10.0\u00b0F (-23.3\u00b0C), we need to determine the mass of ethylene glycol (C\u2082H\u2086O\u2082) to add to 8.5 L of water. The freezing point depression of water is a colligative property, meaning it depends on the number of solute particles in the solution. The formula for freezing point depression is:<\/p>\n\n\n\n<p>\u0394Tf = Kf \u00d7 m<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Tf is the freezing point depression<\/li>\n\n\n\n<li>Kf is the cryoscopic constant of water (1.86\u00b0C\u00b7kg\/mol)<\/li>\n\n\n\n<li>m is the molality of the solution (mol of solute per kg of solvent)<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: Calculate the required freezing point depression<\/strong><\/p>\n\n\n\n<p>The freezing point of pure water is 0\u00b0C. To achieve a freezing point of -23.3\u00b0C, the depression must be:<\/p>\n\n\n\n<p>\u0394Tf = 0\u00b0C &#8211; (-23.3\u00b0C) = 23.3\u00b0C<\/p>\n\n\n\n<p><strong>Step 2: Calculate the molality of the solution<\/strong><\/p>\n\n\n\n<p>Rearranging the freezing point depression formula to solve for molality:<\/p>\n\n\n\n<p>m = \u0394Tf \/ Kf<\/p>\n\n\n\n<p>m = 23.3\u00b0C \/ 1.86\u00b0C\u00b7kg\/mol \u2248 12.53 mol\/kg<\/p>\n\n\n\n<p><strong>Step 3: Calculate the mass of water<\/strong><\/p>\n\n\n\n<p>Assuming the density of water is 1.0 g\/mL, 8.5 L of water has a mass of:<\/p>\n\n\n\n<p>Mass of water = 8.5 L \u00d7 1000 g\/L = 8500 g = 8.5 kg<\/p>\n\n\n\n<p><strong>Step 4: Calculate the moles of ethylene glycol required<\/strong><\/p>\n\n\n\n<p>Moles of ethylene glycol = molality \u00d7 mass of water (in kg)<\/p>\n\n\n\n<p>Moles of ethylene glycol = 12.53 mol\/kg \u00d7 8.5 kg \u2248 106.5 mol<\/p>\n\n\n\n<p><strong>Step 5: Calculate the mass of ethylene glycol required<\/strong><\/p>\n\n\n\n<p>Mass of ethylene glycol = moles \u00d7 molar mass<\/p>\n\n\n\n<p>Mass of ethylene glycol = 106.5 mol \u00d7 62.1 g\/mol \u2248 6620.65 g \u2248 6.62 kg<\/p>\n\n\n\n<p><strong>Conclusion:<\/strong><\/p>\n\n\n\n<p>To prepare an antifreeze solution that freezes at -10.0\u00b0F (-23.3\u00b0C), you need to add approximately 6.62 kg of ethylene glycol to 8.5 kg of water.<\/p>\n\n\n\n<p><strong>Note:<\/strong> This calculation assumes ideal behavior and does not account for deviations from ideality that may occur at high concentrations of ethylene glycol. In practice, the actual amount needed may vary slightly. Additionally, it&#8217;s important to consider the toxicity of ethylene glycol and handle it with care. Always follow safety guidelines when preparing and handling antifreeze solutions.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00c3\u201a\u00c2 What mass of ethylene glycol (C2H6O2, molar mass = 62.1 g\/mol), the main component of antifreeze, must be added to 8.5 L water to produce a solution for use in a car\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2s radiator that freezes at -10.0oF (-23.3oC)? Assume the density of water is exactly 1.0 g\/mL. Kf of water = 1.86 oC\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u0153kg\/mol Molar [&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-182148","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182148","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=182148"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182148\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182148"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182148"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182148"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}