50.0 g of methanol (CH3OH, molecular mass = 32.04 g\/mol) is dissolved to make 500.0 mL of solution. What is the molarity, mol\/L, of the resulting solution?<\/p>\n\n\n\n
Would that be?<\/p>\n\n\n\n
32.04\/.50 = 64.08<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the molarity (mol\/L)<\/strong> of a solution, use the formula: Molarity (M)=moles of soluteliters of solution\\text{Molarity (M)} = \\frac{\\text{moles of solute}}{\\text{liters of solution}}Molarity (M)=liters of solutionmoles of solute\u200b<\/p>\n\n\n\n Given:<\/p>\n\n\n\n Moles of CH3OH=50.0 g32.04 g\/mol\u22481.561 mol\\text{Moles of CH}_3\\text{OH} = \\frac{50.0\\ \\text{g}}{32.04\\ \\text{g\/mol}} \\approx 1.561\\ \\text{mol}Moles of CH3\u200bOH=32.04 g\/mol50.0 g\u200b\u22481.561 mol<\/p>\n\n\n\n Given:<\/p>\n\n\n\n Molarity=1.561 mol0.5000 L=3.122 mol\/L\\text{Molarity} = \\frac{1.561\\ \\text{mol}}{0.5000\\ \\text{L}} = 3.122\\ \\text{mol\/L}Molarity=0.5000 L1.561 mol\u200b=3.122 mol\/L<\/p>\n\n\n\n Molarity is a common unit used in chemistry to express the concentration of a solution. It tells us how many moles of solute are present in one liter of solution. To find the molarity of a methanol solution, we need two pieces of information: how many moles<\/strong> of methanol are dissolved and what the total volume<\/strong> of the solution is in liters<\/strong>.<\/p>\n\n\n\n We start by converting the mass of methanol (50.0 g) into moles using the compound\u2019s molar mass<\/strong>, which represents the mass of one mole of a substance. Methanol (CH\u2083OH) has a molar mass of 32.04 g\/mol, which we obtain by summing the atomic masses of its elements (C: 12.01, H: 1.008 \u00d7 4, O: 16.00). Dividing the given mass by this molar mass yields 1.561 moles.<\/p>\n\n\n\n Next, we convert the solution\u2019s volume from milliliters to liters because molarity is defined per liter. Since 1000 milliliters make one liter, 500.0 mL equals 0.5000 L.<\/p>\n\n\n\n Finally, we divide the moles of methanol by the volume of the solution in liters to get the molarity: 1.561 mol0.5000 L=3.122 mol\/L\\frac{1.561\\ \\text{mol}}{0.5000\\ \\text{L}} = 3.122\\ \\text{mol\/L}0.5000 L1.561 mol\u200b=3.122 mol\/L<\/p>\n\n\n\n Thus, the concentration of the methanol solution is approximately 3.12 mol\/L<\/strong>. This value reflects how densely the methanol is packed into the given volume of solvent, which is crucial for predicting how the solution will behave in chemical reactions.<\/p>\n","protected":false},"excerpt":{"rendered":" 50.0 g of methanol (CH3OH, molecular mass = 32.04 g\/mol) is dissolved to make 500.0 mL of solution. What is the molarity, mol\/L, of the resulting solution? Would that be? 32.04\/.50 = 64.08 The correct answer and explanation is : To calculate the molarity (mol\/L) of a solution, use the formula: Molarity (M)=moles of soluteliters of solution\\text{Molarity (M)} = \\frac{\\text{moles […]<\/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-201256","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201256","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=201256"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201256\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201256"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201256"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201256"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Calculate moles of methanol<\/h3>\n\n\n\n
\n
Step 2: Convert volume of solution to liters<\/h3>\n\n\n\n
\n
Step 3: Calculate molarity<\/h3>\n\n\n\n
\n\n\n\n\u2705 Final Answer: 3.12 mol\/L<\/strong><\/h3>\n\n\n\n
\n\n\n\n\ud83d\udcd8 Explanation (300 words)<\/strong><\/h3>\n\n\n\n