{"id":248331,"date":"2025-07-08T10:08:38","date_gmt":"2025-07-08T10:08:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248331"},"modified":"2025-07-08T10:08:40","modified_gmt":"2025-07-08T10:08:40","slug":"a-solution-of-hsoaq-with-a-molal-concentration-of-2-24-m-has-a-density-of-1-135-g-ml-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/a-solution-of-hsoaq-with-a-molal-concentration-of-2-24-m-has-a-density-of-1-135-g-ml-2\/","title":{"rendered":"A solution of HSO(aq) with a molal concentration of 2.24 m has a density of 1.135 g\/mL."},"content":{"rendered":"\n<p>A solution of H<br>SO<br>(aq) with a molal concentration of 2.24 m has a density of 1.135 g\/mL. What is the molar concentration of this solution? molar concentration:<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the molar concentration of a solution, we need to use the relationship between molality, density, and molar concentration. Here&#8217;s how you can do it step by step:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molality (m) of the solution = 2.24 mol\/kg<\/li>\n\n\n\n<li>Density (\u03c1) of the solution = 1.135 g\/mL<\/li>\n\n\n\n<li>The solvent in this case is water (H\u2082SO\u2084 is dissolved in water).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Convert molality to moles of solute per kilogram of solvent.<\/h3>\n\n\n\n<p>Molality (m) is defined as the number of moles of solute (H\u2082SO\u2084) per kilogram of solvent. So, the molality (2.24 mol\/kg) means there are 2.24 moles of H\u2082SO\u2084 per 1 kilogram of water.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the mass of the solution.<\/h3>\n\n\n\n<p>Density (\u03c1) is given as 1.135 g\/mL. To convert this to kilograms per liter (kg\/L), we multiply by 1000 (since 1 g\/mL = 1000 kg\/L):1.135\u2009g\/mL=1135\u2009g\/L=1.135\u2009kg\/L.1.135 \\, \\text{g\/mL} = 1135 \\, \\text{g\/L} = 1.135 \\, \\text{kg\/L}.1.135g\/mL=1135g\/L=1.135kg\/L.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Find the mass of 1 liter of solution.<\/h3>\n\n\n\n<p>1 liter of solution weighs 1.135 kg.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Use molality to calculate moles of solute in 1 liter of solution.<\/h3>\n\n\n\n<p>Since 2.24 mol of H\u2082SO\u2084 are dissolved in 1 kg of water, and the mass of the solution is 1.135 kg, we can figure out the moles of H\u2082SO\u2084 in 1 liter of solution.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of solvent (water) in 1 liter of solution = 1.135 kg (total solution mass) &#8211; 1 kg (mass of water) = 0.135 kg of water.<\/li>\n\n\n\n<li>The molality gives the amount of H\u2082SO\u2084 per kilogram of water:<\/li>\n<\/ul>\n\n\n\n<p>Moles&nbsp;of&nbsp;H\u2082SO\u2084=2.24\u2009mol\/kg\u00d70.135\u2009kg=0.3024\u2009mol.\\text{Moles of H\u2082SO\u2084} = 2.24 \\, \\text{mol\/kg} \\times 0.135 \\, \\text{kg} = 0.3024 \\, \\text{mol}.Moles&nbsp;of&nbsp;H\u2082SO\u2084=2.24mol\/kg\u00d70.135kg=0.3024mol.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Calculate the molar concentration.<\/h3>\n\n\n\n<p>Molar concentration is the number of moles of solute per liter of solution. Since we have 0.3024 moles of H\u2082SO\u2084 in 1 liter of solution, the molar concentration is:Molar&nbsp;concentration=0.3024\u2009mol\/L.\\text{Molar concentration} = 0.3024 \\, \\text{mol\/L}.Molar&nbsp;concentration=0.3024mol\/L.<\/p>\n\n\n\n<p>Thus, the molar concentration of the solution is <strong>0.302 mol\/L<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>By understanding the relationship between molality, density, and volume, we can determine that the molar concentration of the H\u2082SO\u2084 solution is 0.302 mol\/L.<\/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-banner8-205.jpeg\" alt=\"\" class=\"wp-image-248332\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A solution of HSO(aq) with a molal concentration of 2.24 m has a density of 1.135 g\/mL. What is the molar concentration of this solution? molar concentration: The Correct Answer and Explanation is: To calculate the molar concentration of a solution, we need to use the relationship between molality, density, and molar concentration. Here&#8217;s how [&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-248331","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248331","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=248331"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248331\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248331"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248331"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248331"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}