What is the mole fraction of solute in a 3.68 m aqueous solution?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the mole fraction of the solute in a solution, we need to understand a few concepts:<\/p>\n\n\n\n [ [ [ [ The mole fraction of the solute in the 3.68 m aqueous solution is approximately 0.0622<\/strong>.<\/p>\n\n\n\n This result shows that the solute makes up only a small fraction of the total number of moles in the solution, which is expected for a dilute solution.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the mole fraction of solute in a 3.68 m aqueous solution? The correct answer and explanation is : To calculate the mole fraction of the solute in a solution, we need to understand a few concepts: [X_{\\text{solute}} = \\frac{\\text{moles of solute}}{\\text{moles of solute} + \\text{moles of solvent}}] Steps to find the mole fraction […]<\/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-201327","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201327","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=201327"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201327\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201327"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201327"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201327"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
X_{\\text{solute}} = \\frac{\\text{moles of solute}}{\\text{moles of solute} + \\text{moles of solvent}}
]<\/p>\n\n\n\nSteps to find the mole fraction of solute:<\/h3>\n\n\n\n
\n
The molality is 3.68 m, which means 3.68 moles of solute per 1 kilogram of solvent. So, we have 3.68 moles of solute.<\/li>\n\n\n\n
The solvent is water (H\u2082O). The molar mass of water is approximately 18.015 g\/mol. Since there is 1 kg of solvent, we can convert the mass of solvent to moles:<\/li>\n<\/ol>\n\n\n\n
\\text{moles of solvent} = \\frac{1000 \\, \\text{g}}{18.015 \\, \\text{g\/mol}} = 55.51 \\, \\text{mol}
]<\/p>\n\n\n\n\n
Now, we can calculate the mole fraction of the solute using the formula:<\/li>\n<\/ol>\n\n\n\n
X_{\\text{solute}} = \\frac{\\text{moles of solute}}{\\text{moles of solute} + \\text{moles of solvent}} = \\frac{3.68}{3.68 + 55.51}
]<\/p>\n\n\n\n
X_{\\text{solute}} = \\frac{3.68}{59.19} = 0.0622
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n