{"id":248297,"date":"2025-07-08T09:38:25","date_gmt":"2025-07-08T09:38:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248297"},"modified":"2025-07-08T09:38:27","modified_gmt":"2025-07-08T09:38:27","slug":"a-solution-of-h2so4-with-a-molal-concentration-of-5-25m-has-a-density-of-1-266-g-ml","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/a-solution-of-h2so4-with-a-molal-concentration-of-5-25m-has-a-density-of-1-266-g-ml\/","title":{"rendered":"A solution of H2SO4 with a molal concentration of 5.25m has a density of 1.266 g\/ml"},"content":{"rendered":"\n<p>A solution of H2SO4 with a molal concentration of 5.25m has a density of 1.266 g\/ml. what is the molar concentration of this solution?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the molar concentration of the solution, we first need to understand the relationship between molality and molarity.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Define the given values<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molality (m)<\/strong>: 5.25 mol\/kg<\/li>\n\n\n\n<li><strong>Density (d)<\/strong>: 1.266 g\/mL<\/li>\n\n\n\n<li><strong>Molar mass of H2SO4<\/strong>: 98.079 g\/mol<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Convert the density from g\/mL to g\/L<\/h3>\n\n\n\n<p>Since we are working with density in g\/mL and molarity requires units in mol\/L, we first convert the density:1.266\u2009g\/mL\u00d71000\u2009mL\/L=1266\u2009g\/L1.266 \\, \\text{g\/mL} \\times 1000 \\, \\text{mL\/L} = 1266 \\, \\text{g\/L}1.266g\/mL\u00d71000mL\/L=1266g\/L<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the mass of the solution<\/h3>\n\n\n\n<p>Molality is defined as moles of solute per kilogram of solvent (mol\/kg). To calculate molarity, we need to know the total volume of the solution. Let&#8217;s assume that we have 1 kg of solvent. Since molality is 5.25 mol\/kg, this corresponds to 5.25 moles of H2SO4 in 1 kg of solvent.<\/p>\n\n\n\n<p>Now, the mass of H2SO4 in 5.25 moles is:Mass&nbsp;of&nbsp;H2SO4=5.25\u2009mol\u00d798.079\u2009g\/mol=514.41675\u2009g\\text{Mass of H2SO4} = 5.25 \\, \\text{mol} \\times 98.079 \\, \\text{g\/mol} = 514.41675 \\, \\text{g}Mass&nbsp;of&nbsp;H2SO4=5.25mol\u00d798.079g\/mol=514.41675g<\/p>\n\n\n\n<p>Thus, the total mass of the solution is the mass of the solvent (1000 g) plus the mass of H2SO4 (514.41675 g):Total&nbsp;mass&nbsp;of&nbsp;solution=1000\u2009g+514.41675\u2009g=1514.41675\u2009g\\text{Total mass of solution} = 1000 \\, \\text{g} + 514.41675 \\, \\text{g} = 1514.41675 \\, \\text{g}Total&nbsp;mass&nbsp;of&nbsp;solution=1000g+514.41675g=1514.41675g<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate the volume of the solution<\/h3>\n\n\n\n<p>Using the density formula, we can calculate the volume of the solution:Volume&nbsp;of&nbsp;solution=Mass&nbsp;of&nbsp;solutionDensity=1514.41675\u2009g1266\u2009g\/L=1.195\u2009L\\text{Volume of solution} = \\frac{\\text{Mass of solution}}{\\text{Density}} = \\frac{1514.41675 \\, \\text{g}}{1266 \\, \\text{g\/L}} = 1.195 \\, \\text{L}Volume&nbsp;of&nbsp;solution=DensityMass&nbsp;of&nbsp;solution\u200b=1266g\/L1514.41675g\u200b=1.195L<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Calculate the molarity<\/h3>\n\n\n\n<p>Finally, molarity (M) is defined as the number of moles of solute per liter of solution:Molarity=Moles&nbsp;of&nbsp;H2SO4Volume&nbsp;of&nbsp;solution=5.25\u2009mol1.195\u2009L=4.39\u2009M\\text{Molarity} = \\frac{\\text{Moles of H2SO4}}{\\text{Volume of solution}} = \\frac{5.25 \\, \\text{mol}}{1.195 \\, \\text{L}} = 4.39 \\, \\text{M}Molarity=Volume&nbsp;of&nbsp;solutionMoles&nbsp;of&nbsp;H2SO4\u200b=1.195L5.25mol\u200b=4.39M<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The molar concentration of the H2SO4 solution is <strong>4.39 M<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-197.jpeg\" alt=\"\" class=\"wp-image-248298\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A solution of H2SO4 with a molal concentration of 5.25m has a density of 1.266 g\/ml. what is the molar concentration of this solution? The Correct Answer and Explanation is: To calculate the molar concentration of the solution, we first need to understand the relationship between molality and molarity. Step 1: Define the given values [&hellip;]<\/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-248297","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248297","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=248297"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248297\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}