To solve the problem step-by-step, let’s go over each part.<\/p>\n\n\n\n
1. Mole Fraction of Methanol in the Solution<\/strong><\/h3>\n\n\n\nMole fraction<\/strong> is the ratio of the number of moles of a substance to the total number of moles in the solution. We can calculate the mole fraction of methanol (CH\u2083OH) in the solution using the following steps:<\/p>\n\n\n\nStep 1: Find the mass of CH\u2083OH and CH\u2083CN<\/h4>\n\n\n\n\n- Mass of CH\u2083OH (methanol)<\/strong>:
The volume of methanol is given as 21.5 mL, and its density is 0.791 g\/mL. Mass\u00a0of\u00a0CH\u2083OH=Density\u00a0of\u00a0CH\u2083OH\u00d7Volume\u00a0of\u00a0CH\u2083OH\\text{Mass of CH\u2083OH} = \\text{Density of CH\u2083OH} \\times \\text{Volume of CH\u2083OH}Mass\u00a0of\u00a0CH\u2083OH=Density\u00a0of\u00a0CH\u2083OH\u00d7Volume\u00a0of\u00a0CH\u2083OH Mass\u00a0of\u00a0CH\u2083OH=0.791\u2009g\/mL\u00d721.5\u2009mL=17.0\u2009g\\text{Mass of CH\u2083OH} = 0.791 \\, \\text{g\/mL} \\times 21.5 \\, \\text{mL} = 17.0 \\, \\text{g}Mass\u00a0of\u00a0CH\u2083OH=0.791g\/mL\u00d721.5mL=17.0g<\/li>\n\n\n\n- Mass of CH\u2083CN (acetonitrile)<\/strong>:
The volume of acetonitrile is 98.8 mL, and its density is 0.786 g\/mL. Mass\u00a0of\u00a0CH\u2083CN=0.786\u2009g\/mL\u00d798.8\u2009mL=77.7\u2009g\\text{Mass of CH\u2083CN} = 0.786 \\, \\text{g\/mL} \\times 98.8 \\, \\text{mL} = 77.7 \\, \\text{g}Mass\u00a0of\u00a0CH\u2083CN=0.786g\/mL\u00d798.8mL=77.7g<\/li>\n<\/ul>\n\n\n\nStep 2: Calculate the moles of CH\u2083OH and CH\u2083CN<\/h4>\n\n\n\n\n- Moles of CH\u2083OH<\/strong>:
The molar mass of CH\u2083OH is 32.04 g\/mol. Moles\u00a0of\u00a0CH\u2083OH=Mass\u00a0of\u00a0CH\u2083OHMolar\u00a0mass\u00a0of\u00a0CH\u2083OH=17.0\u2009g32.04\u2009g\/mol=0.531\u2009mol\\text{Moles of CH\u2083OH} = \\frac{\\text{Mass of CH\u2083OH}}{\\text{Molar mass of CH\u2083OH}} = \\frac{17.0 \\, \\text{g}}{32.04 \\, \\text{g\/mol}} = 0.531 \\, \\text{mol}Moles\u00a0of\u00a0CH\u2083OH=Molar\u00a0mass\u00a0of\u00a0CH\u2083OHMass\u00a0of\u00a0CH\u2083OH\u200b=32.04g\/mol17.0g\u200b=0.531mol<\/li>\n\n\n\n- Moles of CH\u2083CN<\/strong>:
The molar mass of CH\u2083CN is 41.05 g\/mol. Moles\u00a0of\u00a0CH\u2083CN=Mass\u00a0of\u00a0CH\u2083CNMolar\u00a0mass\u00a0of\u00a0CH\u2083CN=77.7\u2009g41.05\u2009g\/mol=1.89\u2009mol\\text{Moles of CH\u2083CN} = \\frac{\\text{Mass of CH\u2083CN}}{\\text{Molar mass of CH\u2083CN}} = \\frac{77.7 \\, \\text{g}}{41.05 \\, \\text{g\/mol}} = 1.89 \\, \\text{mol}Moles\u00a0of\u00a0CH\u2083CN=Molar\u00a0mass\u00a0of\u00a0CH\u2083CNMass\u00a0of\u00a0CH\u2083CN\u200b=41.05g\/mol77.7g\u200b=1.89mol<\/li>\n<\/ul>\n\n\n\nStep 3: Calculate the mole fraction of CH\u2083OH<\/h4>\n\n\n\n
Mole fraction of CH\u2083OH=Moles of CH\u2083OHMoles of CH\u2083OH+Moles of CH\u2083CN=0.531\u2009mol0.531\u2009mol+1.89\u2009mol=0.219\\text{Mole fraction of CH\u2083OH} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Moles of CH\u2083OH} + \\text{Moles of CH\u2083CN}} = \\frac{0.531 \\, \\text{mol}}{0.531 \\, \\text{mol} + 1.89 \\, \\text{mol}} = 0.219Mole fraction of CH\u2083OH=Moles of CH\u2083OH+Moles of CH\u2083CNMoles of CH\u2083OH\u200b=0.531mol+1.89mol0.531mol\u200b=0.219<\/p>\n\n\n\n
2. Molality of the Solution<\/strong><\/h3>\n\n\n\nMolality<\/strong> is defined as the number of moles of solute (CH\u2083OH) per kilogram of solvent (CH\u2083CN).<\/p>\n\n\n\nStep 1: Find the mass of CH\u2083CN in kilograms<\/h4>\n\n\n\n
The mass of CH\u2083CN is 77.7 g, which is equivalent to:Mass of CH\u2083CN=77.7\u2009g=0.0777\u2009kg\\text{Mass of CH\u2083CN} = 77.7 \\, \\text{g} = 0.0777 \\, \\text{kg}Mass of CH\u2083CN=77.7g=0.0777kg<\/p>\n\n\n\n
Step 2: Calculate the molality<\/h4>\n\n\n\n
Molality=Moles of CH\u2083OHMass of CH\u2083CN (kg)=0.531\u2009mol0.0777\u2009kg=6.83\u2009mol\/kg\\text{Molality} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Mass of CH\u2083CN (kg)}} = \\frac{0.531 \\, \\text{mol}}{0.0777 \\, \\text{kg}} = 6.83 \\, \\text{mol\/kg}Molality=Mass of CH\u2083CN (kg)Moles of CH\u2083OH\u200b=0.0777kg0.531mol\u200b=6.83mol\/kg<\/p>\n\n\n\n
3. Molarity of CH\u2083OH in the Solution<\/strong><\/h3>\n\n\n\nMolarity<\/strong> is the number of moles of solute (CH\u2083OH) per liter of solution.<\/p>\n\n\n\nStep 1: Calculate the total volume of the solution<\/h4>\n\n\n\n
The total volume is the sum of the volumes of CH\u2083OH and CH\u2083CN:Total volume=21.5\u2009mL+98.8\u2009mL=120.3\u2009mL=0.1203\u2009L\\text{Total volume} = 21.5 \\, \\text{mL} + 98.8 \\, \\text{mL} = 120.3 \\, \\text{mL} = 0.1203 \\, \\text{L}Total volume=21.5mL+98.8mL=120.3mL=0.1203L<\/p>\n\n\n\n
Step 2: Calculate the molarity<\/h4>\n\n\n\n
Molarity=Moles of CH\u2083OHTotal volume of solution (L)=0.531\u2009mol0.1203\u2009L=4.42\u2009M\\text{Molarity} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Total volume of solution (L)}} = \\frac{0.531 \\, \\text{mol}}{0.1203 \\, \\text{L}} = 4.42 \\, \\text{M}Molarity=Total volume of solution (L)Moles of CH\u2083OH\u200b=0.1203L0.531mol\u200b=4.42M<\/p>\n\n\n\n
Summary of Results:<\/h3>\n\n\n\n\n- The mole fraction of methanol (CH\u2083OH) in the solution is 0.219<\/strong>.<\/li>\n\n\n\n
- The molality of the solution is 6.83 mol\/kg<\/strong>.<\/li>\n\n\n\n
- The molarity of CH\u2083OH in the solution is 4.42 M<\/strong>.<\/li>\n<\/ol>\n\n\n\n
This solution involves a detailed process of using the given densities, volumes, and molar masses to find the necessary properties, and the assumption that volumes are additive is critical in determining the molarity.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-663.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"The density of acetonitrile (CH3CN)(CH3CN) is 0.786 g\/mLg\/mL, and the density of methanol (CH3OH)(CH3OH) is 0.791 g\/mLg\/mL. A solution is made by dissolving 21.5 mLmL CH3OHCH3OH in 98.8 mLmL CH3CNCH3CN. 1. What is the mole fraction of methanol in the solution? 2. What is the molality of the solution? Assuming CH3OHCH3OH is the solute and […]<\/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-245573","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573","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=245573"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 1: Find the mass of CH\u2083OH and CH\u2083CN<\/h4>\n\n\n\n\n- Mass of CH\u2083OH (methanol)<\/strong>:
The volume of methanol is given as 21.5 mL, and its density is 0.791 g\/mL. Mass\u00a0of\u00a0CH\u2083OH=Density\u00a0of\u00a0CH\u2083OH\u00d7Volume\u00a0of\u00a0CH\u2083OH\\text{Mass of CH\u2083OH} = \\text{Density of CH\u2083OH} \\times \\text{Volume of CH\u2083OH}Mass\u00a0of\u00a0CH\u2083OH=Density\u00a0of\u00a0CH\u2083OH\u00d7Volume\u00a0of\u00a0CH\u2083OH Mass\u00a0of\u00a0CH\u2083OH=0.791\u2009g\/mL\u00d721.5\u2009mL=17.0\u2009g\\text{Mass of CH\u2083OH} = 0.791 \\, \\text{g\/mL} \\times 21.5 \\, \\text{mL} = 17.0 \\, \\text{g}Mass\u00a0of\u00a0CH\u2083OH=0.791g\/mL\u00d721.5mL=17.0g<\/li>\n\n\n\n- Mass of CH\u2083CN (acetonitrile)<\/strong>:
The volume of acetonitrile is 98.8 mL, and its density is 0.786 g\/mL. Mass\u00a0of\u00a0CH\u2083CN=0.786\u2009g\/mL\u00d798.8\u2009mL=77.7\u2009g\\text{Mass of CH\u2083CN} = 0.786 \\, \\text{g\/mL} \\times 98.8 \\, \\text{mL} = 77.7 \\, \\text{g}Mass\u00a0of\u00a0CH\u2083CN=0.786g\/mL\u00d798.8mL=77.7g<\/li>\n<\/ul>\n\n\n\nStep 2: Calculate the moles of CH\u2083OH and CH\u2083CN<\/h4>\n\n\n\n\n- Moles of CH\u2083OH<\/strong>:
The molar mass of CH\u2083OH is 32.04 g\/mol. Moles\u00a0of\u00a0CH\u2083OH=Mass\u00a0of\u00a0CH\u2083OHMolar\u00a0mass\u00a0of\u00a0CH\u2083OH=17.0\u2009g32.04\u2009g\/mol=0.531\u2009mol\\text{Moles of CH\u2083OH} = \\frac{\\text{Mass of CH\u2083OH}}{\\text{Molar mass of CH\u2083OH}} = \\frac{17.0 \\, \\text{g}}{32.04 \\, \\text{g\/mol}} = 0.531 \\, \\text{mol}Moles\u00a0of\u00a0CH\u2083OH=Molar\u00a0mass\u00a0of\u00a0CH\u2083OHMass\u00a0of\u00a0CH\u2083OH\u200b=32.04g\/mol17.0g\u200b=0.531mol<\/li>\n\n\n\n- Moles of CH\u2083CN<\/strong>:
The molar mass of CH\u2083CN is 41.05 g\/mol. Moles\u00a0of\u00a0CH\u2083CN=Mass\u00a0of\u00a0CH\u2083CNMolar\u00a0mass\u00a0of\u00a0CH\u2083CN=77.7\u2009g41.05\u2009g\/mol=1.89\u2009mol\\text{Moles of CH\u2083CN} = \\frac{\\text{Mass of CH\u2083CN}}{\\text{Molar mass of CH\u2083CN}} = \\frac{77.7 \\, \\text{g}}{41.05 \\, \\text{g\/mol}} = 1.89 \\, \\text{mol}Moles\u00a0of\u00a0CH\u2083CN=Molar\u00a0mass\u00a0of\u00a0CH\u2083CNMass\u00a0of\u00a0CH\u2083CN\u200b=41.05g\/mol77.7g\u200b=1.89mol<\/li>\n<\/ul>\n\n\n\nStep 3: Calculate the mole fraction of CH\u2083OH<\/h4>\n\n\n\n
Mole fraction of CH\u2083OH=Moles of CH\u2083OHMoles of CH\u2083OH+Moles of CH\u2083CN=0.531\u2009mol0.531\u2009mol+1.89\u2009mol=0.219\\text{Mole fraction of CH\u2083OH} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Moles of CH\u2083OH} + \\text{Moles of CH\u2083CN}} = \\frac{0.531 \\, \\text{mol}}{0.531 \\, \\text{mol} + 1.89 \\, \\text{mol}} = 0.219Mole fraction of CH\u2083OH=Moles of CH\u2083OH+Moles of CH\u2083CNMoles of CH\u2083OH\u200b=0.531mol+1.89mol0.531mol\u200b=0.219<\/p>\n\n\n\n
2. Molality of the Solution<\/strong><\/h3>\n\n\n\nMolality<\/strong> is defined as the number of moles of solute (CH\u2083OH) per kilogram of solvent (CH\u2083CN).<\/p>\n\n\n\nStep 1: Find the mass of CH\u2083CN in kilograms<\/h4>\n\n\n\n
The mass of CH\u2083CN is 77.7 g, which is equivalent to:Mass of CH\u2083CN=77.7\u2009g=0.0777\u2009kg\\text{Mass of CH\u2083CN} = 77.7 \\, \\text{g} = 0.0777 \\, \\text{kg}Mass of CH\u2083CN=77.7g=0.0777kg<\/p>\n\n\n\n
Step 2: Calculate the molality<\/h4>\n\n\n\n
Molality=Moles of CH\u2083OHMass of CH\u2083CN (kg)=0.531\u2009mol0.0777\u2009kg=6.83\u2009mol\/kg\\text{Molality} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Mass of CH\u2083CN (kg)}} = \\frac{0.531 \\, \\text{mol}}{0.0777 \\, \\text{kg}} = 6.83 \\, \\text{mol\/kg}Molality=Mass of CH\u2083CN (kg)Moles of CH\u2083OH\u200b=0.0777kg0.531mol\u200b=6.83mol\/kg<\/p>\n\n\n\n
3. Molarity of CH\u2083OH in the Solution<\/strong><\/h3>\n\n\n\nMolarity<\/strong> is the number of moles of solute (CH\u2083OH) per liter of solution.<\/p>\n\n\n\nStep 1: Calculate the total volume of the solution<\/h4>\n\n\n\n
The total volume is the sum of the volumes of CH\u2083OH and CH\u2083CN:Total volume=21.5\u2009mL+98.8\u2009mL=120.3\u2009mL=0.1203\u2009L\\text{Total volume} = 21.5 \\, \\text{mL} + 98.8 \\, \\text{mL} = 120.3 \\, \\text{mL} = 0.1203 \\, \\text{L}Total volume=21.5mL+98.8mL=120.3mL=0.1203L<\/p>\n\n\n\n
Step 2: Calculate the molarity<\/h4>\n\n\n\n
Molarity=Moles of CH\u2083OHTotal volume of solution (L)=0.531\u2009mol0.1203\u2009L=4.42\u2009M\\text{Molarity} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Total volume of solution (L)}} = \\frac{0.531 \\, \\text{mol}}{0.1203 \\, \\text{L}} = 4.42 \\, \\text{M}Molarity=Total volume of solution (L)Moles of CH\u2083OH\u200b=0.1203L0.531mol\u200b=4.42M<\/p>\n\n\n\n
Summary of Results:<\/h3>\n\n\n\n\n- The mole fraction of methanol (CH\u2083OH) in the solution is 0.219<\/strong>.<\/li>\n\n\n\n
- The molality of the solution is 6.83 mol\/kg<\/strong>.<\/li>\n\n\n\n
- The molarity of CH\u2083OH in the solution is 4.42 M<\/strong>.<\/li>\n<\/ol>\n\n\n\n
This solution involves a detailed process of using the given densities, volumes, and molar masses to find the necessary properties, and the assumption that volumes are additive is critical in determining the molarity.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"The density of acetonitrile (CH3CN)(CH3CN) is 0.786 g\/mLg\/mL, and the density of methanol (CH3OH)(CH3OH) is 0.791 g\/mLg\/mL. A solution is made by dissolving 21.5 mLmL CH3OHCH3OH in 98.8 mLmL CH3CNCH3CN. 1. What is the mole fraction of methanol in the solution? 2. What is the molality of the solution? Assuming CH3OHCH3OH is the solute and […]<\/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-245573","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573","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=245573"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
The volume of methanol is given as 21.5 mL, and its density is 0.791 g\/mL. Mass\u00a0of\u00a0CH\u2083OH=Density\u00a0of\u00a0CH\u2083OH\u00d7Volume\u00a0of\u00a0CH\u2083OH\\text{Mass of CH\u2083OH} = \\text{Density of CH\u2083OH} \\times \\text{Volume of CH\u2083OH}Mass\u00a0of\u00a0CH\u2083OH=Density\u00a0of\u00a0CH\u2083OH\u00d7Volume\u00a0of\u00a0CH\u2083OH Mass\u00a0of\u00a0CH\u2083OH=0.791\u2009g\/mL\u00d721.5\u2009mL=17.0\u2009g\\text{Mass of CH\u2083OH} = 0.791 \\, \\text{g\/mL} \\times 21.5 \\, \\text{mL} = 17.0 \\, \\text{g}Mass\u00a0of\u00a0CH\u2083OH=0.791g\/mL\u00d721.5mL=17.0g<\/li>\n\n\n\n
The volume of acetonitrile is 98.8 mL, and its density is 0.786 g\/mL. Mass\u00a0of\u00a0CH\u2083CN=0.786\u2009g\/mL\u00d798.8\u2009mL=77.7\u2009g\\text{Mass of CH\u2083CN} = 0.786 \\, \\text{g\/mL} \\times 98.8 \\, \\text{mL} = 77.7 \\, \\text{g}Mass\u00a0of\u00a0CH\u2083CN=0.786g\/mL\u00d798.8mL=77.7g<\/li>\n<\/ul>\n\n\n\n
Step 2: Calculate the moles of CH\u2083OH and CH\u2083CN<\/h4>\n\n\n\n\n- Moles of CH\u2083OH<\/strong>:
The molar mass of CH\u2083OH is 32.04 g\/mol. Moles\u00a0of\u00a0CH\u2083OH=Mass\u00a0of\u00a0CH\u2083OHMolar\u00a0mass\u00a0of\u00a0CH\u2083OH=17.0\u2009g32.04\u2009g\/mol=0.531\u2009mol\\text{Moles of CH\u2083OH} = \\frac{\\text{Mass of CH\u2083OH}}{\\text{Molar mass of CH\u2083OH}} = \\frac{17.0 \\, \\text{g}}{32.04 \\, \\text{g\/mol}} = 0.531 \\, \\text{mol}Moles\u00a0of\u00a0CH\u2083OH=Molar\u00a0mass\u00a0of\u00a0CH\u2083OHMass\u00a0of\u00a0CH\u2083OH\u200b=32.04g\/mol17.0g\u200b=0.531mol<\/li>\n\n\n\n- Moles of CH\u2083CN<\/strong>:
The molar mass of CH\u2083CN is 41.05 g\/mol. Moles\u00a0of\u00a0CH\u2083CN=Mass\u00a0of\u00a0CH\u2083CNMolar\u00a0mass\u00a0of\u00a0CH\u2083CN=77.7\u2009g41.05\u2009g\/mol=1.89\u2009mol\\text{Moles of CH\u2083CN} = \\frac{\\text{Mass of CH\u2083CN}}{\\text{Molar mass of CH\u2083CN}} = \\frac{77.7 \\, \\text{g}}{41.05 \\, \\text{g\/mol}} = 1.89 \\, \\text{mol}Moles\u00a0of\u00a0CH\u2083CN=Molar\u00a0mass\u00a0of\u00a0CH\u2083CNMass\u00a0of\u00a0CH\u2083CN\u200b=41.05g\/mol77.7g\u200b=1.89mol<\/li>\n<\/ul>\n\n\n\nStep 3: Calculate the mole fraction of CH\u2083OH<\/h4>\n\n\n\n
Mole fraction of CH\u2083OH=Moles of CH\u2083OHMoles of CH\u2083OH+Moles of CH\u2083CN=0.531\u2009mol0.531\u2009mol+1.89\u2009mol=0.219\\text{Mole fraction of CH\u2083OH} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Moles of CH\u2083OH} + \\text{Moles of CH\u2083CN}} = \\frac{0.531 \\, \\text{mol}}{0.531 \\, \\text{mol} + 1.89 \\, \\text{mol}} = 0.219Mole fraction of CH\u2083OH=Moles of CH\u2083OH+Moles of CH\u2083CNMoles of CH\u2083OH\u200b=0.531mol+1.89mol0.531mol\u200b=0.219<\/p>\n\n\n\n
2. Molality of the Solution<\/strong><\/h3>\n\n\n\nMolality<\/strong> is defined as the number of moles of solute (CH\u2083OH) per kilogram of solvent (CH\u2083CN).<\/p>\n\n\n\nStep 1: Find the mass of CH\u2083CN in kilograms<\/h4>\n\n\n\n
The mass of CH\u2083CN is 77.7 g, which is equivalent to:Mass of CH\u2083CN=77.7\u2009g=0.0777\u2009kg\\text{Mass of CH\u2083CN} = 77.7 \\, \\text{g} = 0.0777 \\, \\text{kg}Mass of CH\u2083CN=77.7g=0.0777kg<\/p>\n\n\n\n
Step 2: Calculate the molality<\/h4>\n\n\n\n
Molality=Moles of CH\u2083OHMass of CH\u2083CN (kg)=0.531\u2009mol0.0777\u2009kg=6.83\u2009mol\/kg\\text{Molality} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Mass of CH\u2083CN (kg)}} = \\frac{0.531 \\, \\text{mol}}{0.0777 \\, \\text{kg}} = 6.83 \\, \\text{mol\/kg}Molality=Mass of CH\u2083CN (kg)Moles of CH\u2083OH\u200b=0.0777kg0.531mol\u200b=6.83mol\/kg<\/p>\n\n\n\n
3. Molarity of CH\u2083OH in the Solution<\/strong><\/h3>\n\n\n\nMolarity<\/strong> is the number of moles of solute (CH\u2083OH) per liter of solution.<\/p>\n\n\n\nStep 1: Calculate the total volume of the solution<\/h4>\n\n\n\n
The total volume is the sum of the volumes of CH\u2083OH and CH\u2083CN:Total volume=21.5\u2009mL+98.8\u2009mL=120.3\u2009mL=0.1203\u2009L\\text{Total volume} = 21.5 \\, \\text{mL} + 98.8 \\, \\text{mL} = 120.3 \\, \\text{mL} = 0.1203 \\, \\text{L}Total volume=21.5mL+98.8mL=120.3mL=0.1203L<\/p>\n\n\n\n
Step 2: Calculate the molarity<\/h4>\n\n\n\n
Molarity=Moles of CH\u2083OHTotal volume of solution (L)=0.531\u2009mol0.1203\u2009L=4.42\u2009M\\text{Molarity} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Total volume of solution (L)}} = \\frac{0.531 \\, \\text{mol}}{0.1203 \\, \\text{L}} = 4.42 \\, \\text{M}Molarity=Total volume of solution (L)Moles of CH\u2083OH\u200b=0.1203L0.531mol\u200b=4.42M<\/p>\n\n\n\n
Summary of Results:<\/h3>\n\n\n\n\n- The mole fraction of methanol (CH\u2083OH) in the solution is 0.219<\/strong>.<\/li>\n\n\n\n
- The molality of the solution is 6.83 mol\/kg<\/strong>.<\/li>\n\n\n\n
- The molarity of CH\u2083OH in the solution is 4.42 M<\/strong>.<\/li>\n<\/ol>\n\n\n\n
This solution involves a detailed process of using the given densities, volumes, and molar masses to find the necessary properties, and the assumption that volumes are additive is critical in determining the molarity.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"The density of acetonitrile (CH3CN)(CH3CN) is 0.786 g\/mLg\/mL, and the density of methanol (CH3OH)(CH3OH) is 0.791 g\/mLg\/mL. A solution is made by dissolving 21.5 mLmL CH3OHCH3OH in 98.8 mLmL CH3CNCH3CN. 1. What is the mole fraction of methanol in the solution? 2. What is the molality of the solution? Assuming CH3OHCH3OH is the solute and […]<\/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-245573","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573","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=245573"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
The molar mass of CH\u2083OH is 32.04 g\/mol. Moles\u00a0of\u00a0CH\u2083OH=Mass\u00a0of\u00a0CH\u2083OHMolar\u00a0mass\u00a0of\u00a0CH\u2083OH=17.0\u2009g32.04\u2009g\/mol=0.531\u2009mol\\text{Moles of CH\u2083OH} = \\frac{\\text{Mass of CH\u2083OH}}{\\text{Molar mass of CH\u2083OH}} = \\frac{17.0 \\, \\text{g}}{32.04 \\, \\text{g\/mol}} = 0.531 \\, \\text{mol}Moles\u00a0of\u00a0CH\u2083OH=Molar\u00a0mass\u00a0of\u00a0CH\u2083OHMass\u00a0of\u00a0CH\u2083OH\u200b=32.04g\/mol17.0g\u200b=0.531mol<\/li>\n\n\n\n
The molar mass of CH\u2083CN is 41.05 g\/mol. Moles\u00a0of\u00a0CH\u2083CN=Mass\u00a0of\u00a0CH\u2083CNMolar\u00a0mass\u00a0of\u00a0CH\u2083CN=77.7\u2009g41.05\u2009g\/mol=1.89\u2009mol\\text{Moles of CH\u2083CN} = \\frac{\\text{Mass of CH\u2083CN}}{\\text{Molar mass of CH\u2083CN}} = \\frac{77.7 \\, \\text{g}}{41.05 \\, \\text{g\/mol}} = 1.89 \\, \\text{mol}Moles\u00a0of\u00a0CH\u2083CN=Molar\u00a0mass\u00a0of\u00a0CH\u2083CNMass\u00a0of\u00a0CH\u2083CN\u200b=41.05g\/mol77.7g\u200b=1.89mol<\/li>\n<\/ul>\n\n\n\n
Step 3: Calculate the mole fraction of CH\u2083OH<\/h4>\n\n\n\n
Mole fraction of CH\u2083OH=Moles of CH\u2083OHMoles of CH\u2083OH+Moles of CH\u2083CN=0.531\u2009mol0.531\u2009mol+1.89\u2009mol=0.219\\text{Mole fraction of CH\u2083OH} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Moles of CH\u2083OH} + \\text{Moles of CH\u2083CN}} = \\frac{0.531 \\, \\text{mol}}{0.531 \\, \\text{mol} + 1.89 \\, \\text{mol}} = 0.219Mole fraction of CH\u2083OH=Moles of CH\u2083OH+Moles of CH\u2083CNMoles of CH\u2083OH\u200b=0.531mol+1.89mol0.531mol\u200b=0.219<\/p>\n\n\n\n
2. Molality of the Solution<\/strong><\/h3>\n\n\n\nMolality<\/strong> is defined as the number of moles of solute (CH\u2083OH) per kilogram of solvent (CH\u2083CN).<\/p>\n\n\n\nStep 1: Find the mass of CH\u2083CN in kilograms<\/h4>\n\n\n\n
The mass of CH\u2083CN is 77.7 g, which is equivalent to:Mass of CH\u2083CN=77.7\u2009g=0.0777\u2009kg\\text{Mass of CH\u2083CN} = 77.7 \\, \\text{g} = 0.0777 \\, \\text{kg}Mass of CH\u2083CN=77.7g=0.0777kg<\/p>\n\n\n\n
Step 2: Calculate the molality<\/h4>\n\n\n\n
Molality=Moles of CH\u2083OHMass of CH\u2083CN (kg)=0.531\u2009mol0.0777\u2009kg=6.83\u2009mol\/kg\\text{Molality} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Mass of CH\u2083CN (kg)}} = \\frac{0.531 \\, \\text{mol}}{0.0777 \\, \\text{kg}} = 6.83 \\, \\text{mol\/kg}Molality=Mass of CH\u2083CN (kg)Moles of CH\u2083OH\u200b=0.0777kg0.531mol\u200b=6.83mol\/kg<\/p>\n\n\n\n
3. Molarity of CH\u2083OH in the Solution<\/strong><\/h3>\n\n\n\nMolarity<\/strong> is the number of moles of solute (CH\u2083OH) per liter of solution.<\/p>\n\n\n\nStep 1: Calculate the total volume of the solution<\/h4>\n\n\n\n
The total volume is the sum of the volumes of CH\u2083OH and CH\u2083CN:Total volume=21.5\u2009mL+98.8\u2009mL=120.3\u2009mL=0.1203\u2009L\\text{Total volume} = 21.5 \\, \\text{mL} + 98.8 \\, \\text{mL} = 120.3 \\, \\text{mL} = 0.1203 \\, \\text{L}Total volume=21.5mL+98.8mL=120.3mL=0.1203L<\/p>\n\n\n\n
Step 2: Calculate the molarity<\/h4>\n\n\n\n
Molarity=Moles of CH\u2083OHTotal volume of solution (L)=0.531\u2009mol0.1203\u2009L=4.42\u2009M\\text{Molarity} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Total volume of solution (L)}} = \\frac{0.531 \\, \\text{mol}}{0.1203 \\, \\text{L}} = 4.42 \\, \\text{M}Molarity=Total volume of solution (L)Moles of CH\u2083OH\u200b=0.1203L0.531mol\u200b=4.42M<\/p>\n\n\n\n
Summary of Results:<\/h3>\n\n\n\n\n- The mole fraction of methanol (CH\u2083OH) in the solution is 0.219<\/strong>.<\/li>\n\n\n\n
- The molality of the solution is 6.83 mol\/kg<\/strong>.<\/li>\n\n\n\n
- The molarity of CH\u2083OH in the solution is 4.42 M<\/strong>.<\/li>\n<\/ol>\n\n\n\n
This solution involves a detailed process of using the given densities, volumes, and molar masses to find the necessary properties, and the assumption that volumes are additive is critical in determining the molarity.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"The density of acetonitrile (CH3CN)(CH3CN) is 0.786 g\/mLg\/mL, and the density of methanol (CH3OH)(CH3OH) is 0.791 g\/mLg\/mL. A solution is made by dissolving 21.5 mLmL CH3OHCH3OH in 98.8 mLmL CH3CNCH3CN. 1. What is the mole fraction of methanol in the solution? 2. What is the molality of the solution? Assuming CH3OHCH3OH is the solute and […]<\/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-245573","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573","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=245573"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 1: Find the mass of CH\u2083CN in kilograms<\/h4>\n\n\n\n
The mass of CH\u2083CN is 77.7 g, which is equivalent to:Mass of CH\u2083CN=77.7\u2009g=0.0777\u2009kg\\text{Mass of CH\u2083CN} = 77.7 \\, \\text{g} = 0.0777 \\, \\text{kg}Mass of CH\u2083CN=77.7g=0.0777kg<\/p>\n\n\n\n
Step 2: Calculate the molality<\/h4>\n\n\n\n
Molality=Moles of CH\u2083OHMass of CH\u2083CN (kg)=0.531\u2009mol0.0777\u2009kg=6.83\u2009mol\/kg\\text{Molality} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Mass of CH\u2083CN (kg)}} = \\frac{0.531 \\, \\text{mol}}{0.0777 \\, \\text{kg}} = 6.83 \\, \\text{mol\/kg}Molality=Mass of CH\u2083CN (kg)Moles of CH\u2083OH\u200b=0.0777kg0.531mol\u200b=6.83mol\/kg<\/p>\n\n\n\n
3. Molarity of CH\u2083OH in the Solution<\/strong><\/h3>\n\n\n\nMolarity<\/strong> is the number of moles of solute (CH\u2083OH) per liter of solution.<\/p>\n\n\n\nStep 1: Calculate the total volume of the solution<\/h4>\n\n\n\n
The total volume is the sum of the volumes of CH\u2083OH and CH\u2083CN:Total volume=21.5\u2009mL+98.8\u2009mL=120.3\u2009mL=0.1203\u2009L\\text{Total volume} = 21.5 \\, \\text{mL} + 98.8 \\, \\text{mL} = 120.3 \\, \\text{mL} = 0.1203 \\, \\text{L}Total volume=21.5mL+98.8mL=120.3mL=0.1203L<\/p>\n\n\n\n
Step 2: Calculate the molarity<\/h4>\n\n\n\n
Molarity=Moles of CH\u2083OHTotal volume of solution (L)=0.531\u2009mol0.1203\u2009L=4.42\u2009M\\text{Molarity} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Total volume of solution (L)}} = \\frac{0.531 \\, \\text{mol}}{0.1203 \\, \\text{L}} = 4.42 \\, \\text{M}Molarity=Total volume of solution (L)Moles of CH\u2083OH\u200b=0.1203L0.531mol\u200b=4.42M<\/p>\n\n\n\n
Summary of Results:<\/h3>\n\n\n\n\n- The mole fraction of methanol (CH\u2083OH) in the solution is 0.219<\/strong>.<\/li>\n\n\n\n
- The molality of the solution is 6.83 mol\/kg<\/strong>.<\/li>\n\n\n\n
- The molarity of CH\u2083OH in the solution is 4.42 M<\/strong>.<\/li>\n<\/ol>\n\n\n\n
This solution involves a detailed process of using the given densities, volumes, and molar masses to find the necessary properties, and the assumption that volumes are additive is critical in determining the molarity.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"The density of acetonitrile (CH3CN)(CH3CN) is 0.786 g\/mLg\/mL, and the density of methanol (CH3OH)(CH3OH) is 0.791 g\/mLg\/mL. A solution is made by dissolving 21.5 mLmL CH3OHCH3OH in 98.8 mLmL CH3CNCH3CN. 1. What is the mole fraction of methanol in the solution? 2. What is the molality of the solution? Assuming CH3OHCH3OH is the solute and […]<\/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-245573","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573","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=245573"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 1: Calculate the total volume of the solution<\/h4>\n\n\n\n
The total volume is the sum of the volumes of CH\u2083OH and CH\u2083CN:Total volume=21.5\u2009mL+98.8\u2009mL=120.3\u2009mL=0.1203\u2009L\\text{Total volume} = 21.5 \\, \\text{mL} + 98.8 \\, \\text{mL} = 120.3 \\, \\text{mL} = 0.1203 \\, \\text{L}Total volume=21.5mL+98.8mL=120.3mL=0.1203L<\/p>\n\n\n\n
Step 2: Calculate the molarity<\/h4>\n\n\n\n
Molarity=Moles of CH\u2083OHTotal volume of solution (L)=0.531\u2009mol0.1203\u2009L=4.42\u2009M\\text{Molarity} = \\frac{\\text{Moles of CH\u2083OH}}{\\text{Total volume of solution (L)}} = \\frac{0.531 \\, \\text{mol}}{0.1203 \\, \\text{L}} = 4.42 \\, \\text{M}Molarity=Total volume of solution (L)Moles of CH\u2083OH\u200b=0.1203L0.531mol\u200b=4.42M<\/p>\n\n\n\n
Summary of Results:<\/h3>\n\n\n\n\n- The mole fraction of methanol (CH\u2083OH) in the solution is 0.219<\/strong>.<\/li>\n\n\n\n
- The molality of the solution is 6.83 mol\/kg<\/strong>.<\/li>\n\n\n\n
- The molarity of CH\u2083OH in the solution is 4.42 M<\/strong>.<\/li>\n<\/ol>\n\n\n\n
This solution involves a detailed process of using the given densities, volumes, and molar masses to find the necessary properties, and the assumption that volumes are additive is critical in determining the molarity.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"The density of acetonitrile (CH3CN)(CH3CN) is 0.786 g\/mLg\/mL, and the density of methanol (CH3OH)(CH3OH) is 0.791 g\/mLg\/mL. A solution is made by dissolving 21.5 mLmL CH3OHCH3OH in 98.8 mLmL CH3CNCH3CN. 1. What is the mole fraction of methanol in the solution? 2. What is the molality of the solution? Assuming CH3OHCH3OH is the solute and […]<\/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-245573","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573","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=245573"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
This solution involves a detailed process of using the given densities, volumes, and molar masses to find the necessary properties, and the assumption that volumes are additive is critical in determining the molarity.<\/p>\n\n\n\n The density of acetonitrile (CH3CN)(CH3CN) is 0.786 g\/mLg\/mL, and the density of methanol (CH3OH)(CH3OH) is 0.791 g\/mLg\/mL. A solution is made by dissolving 21.5 mLmL CH3OHCH3OH in 98.8 mLmL CH3CNCH3CN. 1. What is the mole fraction of methanol in the solution? 2. What is the molality of the solution? Assuming CH3OHCH3OH is the solute and […]<\/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-245573","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573","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=245573"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245573\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}<\/figure>\n","protected":false},"excerpt":{"rendered":"