{"id":198153,"date":"2025-03-08T12:32:38","date_gmt":"2025-03-08T12:32:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=198153"},"modified":"2025-03-08T12:32:40","modified_gmt":"2025-03-08T12:32:40","slug":"calculate-the-boiling-point-of-a-solution-of-500-0-g-of-ethylene-glycol-c2h6o2-dissolved-in-500-0-g-of-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/08\/calculate-the-boiling-point-of-a-solution-of-500-0-g-of-ethylene-glycol-c2h6o2-dissolved-in-500-0-g-of-water\/","title":{"rendered":"Calculate the boiling point of a solution of 500.0 g of ethylene glycol (C2H6O2) dissolved in 500.0 g of water"},"content":{"rendered":"\n<p>Calculate the boiling point of a solution of 500.0 g of ethylene glycol (C2H6O2) dissolved in 500.0 g of water. Kf = 1.86\u00b0C\/m and Kb = 0.512\u00b0C\/m. Use 100\u00b0C as the boiling point of<\/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 calculate the boiling point of the solution, we need to apply the concept of <strong>boiling point elevation<\/strong>, which is the phenomenon where the boiling point of a solvent is raised when a non-volatile solute is added. The formula for boiling point elevation is:<\/p>\n\n\n\n<p>[<br>\\Delta T_b = K_b \\times m<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(\\Delta T_b) is the change in boiling point.<\/li>\n\n\n\n<li>(K_b) is the ebullioscopic constant (boiling point elevation constant) for the solvent, in this case, water.<\/li>\n\n\n\n<li>(m) is the molality of the solution.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate molality ((m))<\/h3>\n\n\n\n<p>Molality is defined as the number of moles of solute per kilogram of solvent. First, we need to determine the number of moles of ethylene glycol (C\u2082H\u2086O\u2082).<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Molar mass of ethylene glycol (C\u2082H\u2086O\u2082)<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) has an atomic mass of 12.01 g\/mol.<\/li>\n\n\n\n<li>Hydrogen (H) has an atomic mass of 1.008 g\/mol.<\/li>\n\n\n\n<li>Oxygen (O) has an atomic mass of 16.00 g\/mol.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the molar mass of C\u2082H\u2086O\u2082 is:<\/p>\n\n\n\n<p>[<br>M_{\\text{C}_2\\text{H}_6\\text{O}_2} = 2(12.01) + 6(1.008) + 2(16.00) = 62.068 \\, \\text{g\/mol}<br>]<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Moles of ethylene glycol<\/strong>:<\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{moles of C}_2\\text{H}_6\\text{O}_2 = \\frac{500.0 \\, \\text{g}}{62.068 \\, \\text{g\/mol}} = 8.05 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li><strong>Molality ((m))<\/strong>:<\/li>\n<\/ol>\n\n\n\n<p>Molality is the number of moles of solute per kilogram of solvent. Since we have 500.0 g (0.5000 kg) of water:<\/p>\n\n\n\n<p>[<br>m = \\frac{\\text{moles of C}_2\\text{H}_6\\text{O}_2}{\\text{kg of solvent}} = \\frac{8.05 \\, \\text{mol}}{0.5000 \\, \\text{kg}} = 16.10 \\, \\text{mol\/kg}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate boiling point elevation ((\\Delta T_b))<\/h3>\n\n\n\n<p>Now we can use the formula for boiling point elevation:<\/p>\n\n\n\n<p>[<br>\\Delta T_b = K_b \\times m = 0.512 \\, \\text{\u00b0C\/m} \\times 16.10 \\, \\text{mol\/kg} = 8.23 \\, \\text{\u00b0C}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Determine the new boiling point<\/h3>\n\n\n\n<p>The normal boiling point of water is 100\u00b0C, so the new boiling point is:<\/p>\n\n\n\n<p>[<br>\\text{New boiling point} = 100\u00b0C + 8.23\u00b0C = 108.23\u00b0C<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>The boiling point of the solution is <strong>108.23\u00b0C<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-632-1024x677.png\" alt=\"\" class=\"wp-image-198154\"\/><\/figure>\n\n\n\n<p>Here is the visual representation of the boiling point elevation. The bar on the left represents the boiling point of pure water (100\u00b0C), while the bar on the right represents the boiling point of the solution (108.23\u00b0C). The solution&#8217;s boiling point has increased due to the presence of ethylene glycol, as calculated earlier.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the boiling point of a solution of 500.0 g of ethylene glycol (C2H6O2) dissolved in 500.0 g of water. Kf = 1.86\u00b0C\/m and Kb = 0.512\u00b0C\/m. Use 100\u00b0C as the boiling point of The correct answer and explanation is : To calculate the boiling point of the solution, we need to apply the concept [&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-198153","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198153","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=198153"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198153\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=198153"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=198153"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=198153"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}