Given:<\/strong><\/h3>\n\n\n\n\n- Mass of water (solvent), msolvent=99.95\u2009g=0.09995\u2009kgm_{\\text{solvent}} = 99.95 \\, \\text{g} = 0.09995 \\, \\text{kg}msolvent\u200b=99.95g=0.09995kg<\/li>\n\n\n\n
- Mass of ethylene glycol (solute), msolute=55.7\u2009gm_{\\text{solute}} = 55.7 \\, \\text{g}msolute\u200b=55.7g<\/li>\n\n\n\n
- Boiling point of solution = 104.6\u202f\u00b0C<\/li>\n\n\n\n
- Boiling point of pure water = 100.0\u202f\u00b0C<\/li>\n\n\n\n
- Boiling point elevation, \u0394Tb=104.6\u2212100.0=4.6\u2218C\\Delta T_b = 104.6 – 100.0 = 4.6^\\circ \\text{C}\u0394Tb\u200b=104.6\u2212100.0=4.6\u2218C<\/li>\n\n\n\n
- Boiling point elevation constant of water<\/strong>, Kb=0.512\u2009\u00b0C\\cdotpkg\/molK_b = 0.512 \\, \\text{\u00b0C\u00b7kg\/mol}Kb\u200b=0.512\u00b0C\\cdotpkg\/mol<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 1: Use the boiling point elevation formula<\/strong><\/h3>\n\n\n\n\u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m<\/p>\n\n\n\n
Where:<\/p>\n\n\n\n
\n- \u0394Tb\\Delta T_b\u0394Tb\u200b = boiling point elevation<\/li>\n\n\n\n
- iii = van\u2019t Hoff factor (for ethylene glycol, a non-electrolyte, i=1i = 1i=1)<\/li>\n\n\n\n
- KbK_bKb\u200b = ebullioscopic constant<\/li>\n\n\n\n
- mmm = molality of the solution (mol of solute per kg of solvent)<\/li>\n<\/ul>\n\n\n\n
4.6=1\u22c50.512\u22c5(mol solute0.09995)4.6 = 1 \\cdot 0.512 \\cdot \\left( \\frac{\\text{mol solute}}{0.09995} \\right)4.6=1\u22c50.512\u22c5(0.09995mol solute\u200b)mol solute0.09995=4.60.512\u22488.984\\frac{\\text{mol solute}}{0.09995} = \\frac{4.6}{0.512} \\approx 8.9840.09995mol solute\u200b=0.5124.6\u200b\u22488.984mol solute=8.984\u00d70.09995\u22480.897\u2009mol\\text{mol solute} = 8.984 \\times 0.09995 \\approx 0.897 \\, \\text{mol}mol solute=8.984\u00d70.09995\u22480.897mol<\/p>\n\n\n\n
\n\n\n\nStep 2: Find molar mass<\/strong><\/h3>\n\n\n\nMmol=mass of solutemol solute=55.70.897\u224862.1\u2009g\/molM_{\\text{mol}} = \\frac{\\text{mass of solute}}{\\text{mol solute}} = \\frac{55.7}{0.897} \\approx \\boxed{62.1 \\, \\text{g\/mol}}Mmol\u200b=mol solutemass of solute\u200b=0.89755.7\u200b\u224862.1g\/mol\u200b<\/p>\n\n\n\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\nTo determine the molar mass of a solute using boiling point elevation, we rely on colligative properties\u2014those that depend solely on the number of solute particles, not their identity. Ethylene glycol, a common antifreeze, is a non-electrolyte and does not dissociate in solution, so it has a van\u2019t Hoff factor (iii) of 1.<\/p>\n\n\n\n
The boiling point of a solution is higher than that of the pure solvent. This increase in boiling point (\u0394Tb\\Delta T_b\u0394Tb\u200b) is directly proportional to the molality of the solution (moles of solute per kilogram of solvent). The relationship is described by the equation:\u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m<\/p>\n\n\n\n
In our case, the boiling point of the solution is 104.6\u202f\u00b0C, which is 4.6\u202f\u00b0C higher than the boiling point of pure water. Given that the boiling point elevation constant for water (KbK_bKb\u200b) is 0.512\u202f\u00b0C\u00b7kg\/mol and the solvent is water weighing 99.95 g (or 0.09995 kg), we substitute into the equation and solve for the molality. This gives us the number of moles of ethylene glycol dissolved.<\/p>\n\n\n\n
With the moles calculated (approximately 0.897 mol) and the known mass of the solute (55.7 g), we calculate the molar mass using the formula:Molar mass=massmoles\\text{Molar mass} = \\frac{\\text{mass}}{\\text{moles}}Molar mass=molesmass\u200b<\/p>\n\n\n\n
This yields a molar mass of approximately 62.1 g\/mol, which closely matches the known molar mass of ethylene glycol (C\u2082H\u2086O\u2082), confirming the accuracy of the calculation.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-432.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Determination of the molar mass of ethylene glycol Weight of water: 99.95 g Weight of added ethylene glycol: 55.7 g Boiling Point of Solution: 104.6 \u00c2\u00b0C Molar mass of ethylene glycol: Mmol The Correct Answer and Explanation is: Given: Step 1: Use the boiling point elevation formula \u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m […]<\/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-233253","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233253","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=233253"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233253\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Step 1: Use the boiling point elevation formula<\/strong><\/h3>\n\n\n\n\u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m<\/p>\n\n\n\n
Where:<\/p>\n\n\n\n
\n- \u0394Tb\\Delta T_b\u0394Tb\u200b = boiling point elevation<\/li>\n\n\n\n
- iii = van\u2019t Hoff factor (for ethylene glycol, a non-electrolyte, i=1i = 1i=1)<\/li>\n\n\n\n
- KbK_bKb\u200b = ebullioscopic constant<\/li>\n\n\n\n
- mmm = molality of the solution (mol of solute per kg of solvent)<\/li>\n<\/ul>\n\n\n\n
4.6=1\u22c50.512\u22c5(mol solute0.09995)4.6 = 1 \\cdot 0.512 \\cdot \\left( \\frac{\\text{mol solute}}{0.09995} \\right)4.6=1\u22c50.512\u22c5(0.09995mol solute\u200b)mol solute0.09995=4.60.512\u22488.984\\frac{\\text{mol solute}}{0.09995} = \\frac{4.6}{0.512} \\approx 8.9840.09995mol solute\u200b=0.5124.6\u200b\u22488.984mol solute=8.984\u00d70.09995\u22480.897\u2009mol\\text{mol solute} = 8.984 \\times 0.09995 \\approx 0.897 \\, \\text{mol}mol solute=8.984\u00d70.09995\u22480.897mol<\/p>\n\n\n\n
\n\n\n\nStep 2: Find molar mass<\/strong><\/h3>\n\n\n\nMmol=mass of solutemol solute=55.70.897\u224862.1\u2009g\/molM_{\\text{mol}} = \\frac{\\text{mass of solute}}{\\text{mol solute}} = \\frac{55.7}{0.897} \\approx \\boxed{62.1 \\, \\text{g\/mol}}Mmol\u200b=mol solutemass of solute\u200b=0.89755.7\u200b\u224862.1g\/mol\u200b<\/p>\n\n\n\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\nTo determine the molar mass of a solute using boiling point elevation, we rely on colligative properties\u2014those that depend solely on the number of solute particles, not their identity. Ethylene glycol, a common antifreeze, is a non-electrolyte and does not dissociate in solution, so it has a van\u2019t Hoff factor (iii) of 1.<\/p>\n\n\n\n
The boiling point of a solution is higher than that of the pure solvent. This increase in boiling point (\u0394Tb\\Delta T_b\u0394Tb\u200b) is directly proportional to the molality of the solution (moles of solute per kilogram of solvent). The relationship is described by the equation:\u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m<\/p>\n\n\n\n
In our case, the boiling point of the solution is 104.6\u202f\u00b0C, which is 4.6\u202f\u00b0C higher than the boiling point of pure water. Given that the boiling point elevation constant for water (KbK_bKb\u200b) is 0.512\u202f\u00b0C\u00b7kg\/mol and the solvent is water weighing 99.95 g (or 0.09995 kg), we substitute into the equation and solve for the molality. This gives us the number of moles of ethylene glycol dissolved.<\/p>\n\n\n\n
With the moles calculated (approximately 0.897 mol) and the known mass of the solute (55.7 g), we calculate the molar mass using the formula:Molar mass=massmoles\\text{Molar mass} = \\frac{\\text{mass}}{\\text{moles}}Molar mass=molesmass\u200b<\/p>\n\n\n\n
This yields a molar mass of approximately 62.1 g\/mol, which closely matches the known molar mass of ethylene glycol (C\u2082H\u2086O\u2082), confirming the accuracy of the calculation.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Determination of the molar mass of ethylene glycol Weight of water: 99.95 g Weight of added ethylene glycol: 55.7 g Boiling Point of Solution: 104.6 \u00c2\u00b0C Molar mass of ethylene glycol: Mmol The Correct Answer and Explanation is: Given: Step 1: Use the boiling point elevation formula \u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m […]<\/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-233253","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233253","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=233253"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233253\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
4.6=1\u22c50.512\u22c5(mol solute0.09995)4.6 = 1 \\cdot 0.512 \\cdot \\left( \\frac{\\text{mol solute}}{0.09995} \\right)4.6=1\u22c50.512\u22c5(0.09995mol solute\u200b)mol solute0.09995=4.60.512\u22488.984\\frac{\\text{mol solute}}{0.09995} = \\frac{4.6}{0.512} \\approx 8.9840.09995mol solute\u200b=0.5124.6\u200b\u22488.984mol solute=8.984\u00d70.09995\u22480.897\u2009mol\\text{mol solute} = 8.984 \\times 0.09995 \\approx 0.897 \\, \\text{mol}mol solute=8.984\u00d70.09995\u22480.897mol<\/p>\n\n\n\n
\n\n\n\n
Step 2: Find molar mass<\/strong><\/h3>\n\n\n\nMmol=mass of solutemol solute=55.70.897\u224862.1\u2009g\/molM_{\\text{mol}} = \\frac{\\text{mass of solute}}{\\text{mol solute}} = \\frac{55.7}{0.897} \\approx \\boxed{62.1 \\, \\text{g\/mol}}Mmol\u200b=mol solutemass of solute\u200b=0.89755.7\u200b\u224862.1g\/mol\u200b<\/p>\n\n\n\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\nTo determine the molar mass of a solute using boiling point elevation, we rely on colligative properties\u2014those that depend solely on the number of solute particles, not their identity. Ethylene glycol, a common antifreeze, is a non-electrolyte and does not dissociate in solution, so it has a van\u2019t Hoff factor (iii) of 1.<\/p>\n\n\n\n
The boiling point of a solution is higher than that of the pure solvent. This increase in boiling point (\u0394Tb\\Delta T_b\u0394Tb\u200b) is directly proportional to the molality of the solution (moles of solute per kilogram of solvent). The relationship is described by the equation:\u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m<\/p>\n\n\n\n
In our case, the boiling point of the solution is 104.6\u202f\u00b0C, which is 4.6\u202f\u00b0C higher than the boiling point of pure water. Given that the boiling point elevation constant for water (KbK_bKb\u200b) is 0.512\u202f\u00b0C\u00b7kg\/mol and the solvent is water weighing 99.95 g (or 0.09995 kg), we substitute into the equation and solve for the molality. This gives us the number of moles of ethylene glycol dissolved.<\/p>\n\n\n\n
With the moles calculated (approximately 0.897 mol) and the known mass of the solute (55.7 g), we calculate the molar mass using the formula:Molar mass=massmoles\\text{Molar mass} = \\frac{\\text{mass}}{\\text{moles}}Molar mass=molesmass\u200b<\/p>\n\n\n\n
This yields a molar mass of approximately 62.1 g\/mol, which closely matches the known molar mass of ethylene glycol (C\u2082H\u2086O\u2082), confirming the accuracy of the calculation.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Determination of the molar mass of ethylene glycol Weight of water: 99.95 g Weight of added ethylene glycol: 55.7 g Boiling Point of Solution: 104.6 \u00c2\u00b0C Molar mass of ethylene glycol: Mmol The Correct Answer and Explanation is: Given: Step 1: Use the boiling point elevation formula \u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m […]<\/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-233253","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233253","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=233253"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233253\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
To determine the molar mass of a solute using boiling point elevation, we rely on colligative properties\u2014those that depend solely on the number of solute particles, not their identity. Ethylene glycol, a common antifreeze, is a non-electrolyte and does not dissociate in solution, so it has a van\u2019t Hoff factor (iii) of 1.<\/p>\n\n\n\n
The boiling point of a solution is higher than that of the pure solvent. This increase in boiling point (\u0394Tb\\Delta T_b\u0394Tb\u200b) is directly proportional to the molality of the solution (moles of solute per kilogram of solvent). The relationship is described by the equation:\u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m<\/p>\n\n\n\n
In our case, the boiling point of the solution is 104.6\u202f\u00b0C, which is 4.6\u202f\u00b0C higher than the boiling point of pure water. Given that the boiling point elevation constant for water (KbK_bKb\u200b) is 0.512\u202f\u00b0C\u00b7kg\/mol and the solvent is water weighing 99.95 g (or 0.09995 kg), we substitute into the equation and solve for the molality. This gives us the number of moles of ethylene glycol dissolved.<\/p>\n\n\n\n
With the moles calculated (approximately 0.897 mol) and the known mass of the solute (55.7 g), we calculate the molar mass using the formula:Molar mass=massmoles\\text{Molar mass} = \\frac{\\text{mass}}{\\text{moles}}Molar mass=molesmass\u200b<\/p>\n\n\n\n
This yields a molar mass of approximately 62.1 g\/mol, which closely matches the known molar mass of ethylene glycol (C\u2082H\u2086O\u2082), confirming the accuracy of the calculation.<\/p>\n\n\n\n Determination of the molar mass of ethylene glycol Weight of water: 99.95 g Weight of added ethylene glycol: 55.7 g Boiling Point of Solution: 104.6 \u00c2\u00b0C Molar mass of ethylene glycol: Mmol The Correct Answer and Explanation is: Given: Step 1: Use the boiling point elevation formula \u0394Tb=i\u22c5Kb\u22c5m\\Delta T_b = i \\cdot K_b \\cdot m\u0394Tb\u200b=i\u22c5Kb\u200b\u22c5m […]<\/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-233253","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233253","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=233253"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233253\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}<\/figure>\n","protected":false},"excerpt":{"rendered":"