The density of acetonitrile (CH3CN) is 0.786 g\/mL and the density of methanol (CH3OH) is 0.791 g\/mL. A solution is made by dissolving 15.7 mL of CH3OH in 92.3 mL of CH3CN. What is the mole fraction of the methano in this solution?<\/p>\n\n\n\n
The Correct Answer and Explanation is <\/strong><\/mark><\/p>\n\n\n\n The mole fraction of methanol in the solution is 0.180<\/strong> (rounded to three significant figures).<\/p>\n\n\n\n Thus, the mole fraction of methanol is 0.180<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" The density of acetonitrile (CH3CN) is 0.786 g\/mL and the density of methanol (CH3OH) is 0.791 g\/mL. A solution is made by dissolving 15.7 mL of CH3OH in 92.3 mL of CH3CN. What is the mole fraction of the methano in this solution? The Correct Answer and Explanation is The mole fraction of methanol in […]<\/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-191015","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191015","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=191015"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191015\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191015"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191015"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191015"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
\n
We are given a solution of methanol ((\\text{CH}_3\\text{OH})) and acetonitrile ((\\text{CH}_3\\text{CN})), with known volumes and densities. The goal is to find the mole fraction of methanol in the solution.<\/li>\n\n\n\n
Using the density formula:
[
\\text{Mass} = \\text{Density} \\times \\text{Volume}
]<\/li>\n<\/ol>\n\n\n\n\n
[
15.7 \\text{ mL} \\times 0.791 \\text{ g\/mL} = 12.42 \\text{ g}
]<\/li>\n\n\n\n
[
92.3 \\text{ mL} \\times 0.786 \\text{ g\/mL} = 72.57 \\text{ g}
]<\/li>\n<\/ul>\n\n\n\n\n
Using the formula:
[
\\text{Moles} = \\frac{\\text{Mass}}{\\text{Molar Mass}}
]<\/li>\n<\/ol>\n\n\n\n\n
[
\\frac{12.42 \\text{ g}}{32.04 \\text{ g\/mol}} = 0.3877 \\text{ moles}
]<\/li>\n\n\n\n
[
\\frac{72.57 \\text{ g}}{41.05 \\text{ g\/mol}} = 1.768 \\text{ moles}
]<\/li>\n<\/ul>\n\n\n\n\n
The mole fraction is given by:
[
X_{\\text{methanol}} = \\frac{\\text{Moles of methanol}}{\\text{Total moles}}
]
[
X_{\\text{methanol}} = \\frac{0.3877}{0.3877 + 1.768} = \\frac{0.3877}{2.1557} = 0.180
]<\/li>\n<\/ol>\n\n\n\n