To find the molar concentration of the sulfuric acid (H\u2082SO\u2084) solution, we need to use the information given:<\/p>\n\n\n\n
- \n
- Molality (m) = 2.24 mol\/kg<\/li>\n\n\n\n
- Density (\u03c1) = 1.135 g\/mL<\/li>\n\n\n\n
- Molar mass of H\u2082SO\u2084 = 98.08 g\/mol<\/li>\n<\/ul>\n\n\n\n
Step-by-Step Solution:<\/h3>\n\n\n\n
- \n
- Convert molality to moles of H\u2082SO\u2084<\/strong>:
Molality (m) is defined as the number of moles of solute (H\u2082SO\u2084) per kilogram of solvent. Since we have a 2.24 molal solution, this means there are 2.24 moles of H\u2082SO\u2084 in every 1 kilogram (1000 grams) of solvent.<\/li>\n\n\n\n- Find the mass of the solution<\/strong>:
The density of the solution is given as 1.135 g\/mL, which means there are 1.135 grams of solution per 1 milliliter of solution. To calculate the mass of 1 liter (1000 mL) of solution, we multiply: Mass\u00a0of\u00a01\u00a0liter\u00a0of\u00a0solution=1.135\u2009g\/mL\u00d71000\u2009mL=1135\u2009g\\text{Mass of 1 liter of solution} = 1.135 \\, \\text{g\/mL} \\times 1000 \\, \\text{mL} = 1135 \\, \\text{g}Mass\u00a0of\u00a01\u00a0liter\u00a0of\u00a0solution=1.135g\/mL\u00d71000mL=1135g So, the mass of 1 liter of solution is 1135 grams.<\/li>\n\n\n\n- Find the mass of solvent (water)<\/strong>:
Since the solution contains H\u2082SO\u2084 dissolved in water, the mass of the solvent (water) is the mass of the solution minus the mass of the H\u2082SO\u2084. First, find the mass of H\u2082SO\u2084: Mass\u00a0of\u00a0H\u2082SO\u2084=molality\u00d7mass\u00a0of\u00a0solvent=2.24\u2009mol\/kg\u00d71\u2009kg=2.24\u2009mol\\text{Mass of H\u2082SO\u2084} = \\text{molality} \\times \\text{mass of solvent} = 2.24 \\, \\text{mol\/kg} \\times 1 \\, \\text{kg} = 2.24 \\, \\text{mol} Mass\u00a0of\u00a0H\u2082SO\u2084=molality\u00d7mass\u00a0of\u00a0solvent=2.24mol\/kg\u00d71kg=2.24mol Now, multiply by the molar mass of H\u2082SO\u2084 to get the mass of H\u2082SO\u2084: Mass\u00a0of\u00a0H\u2082SO\u2084=2.24\u2009mol\u00d798.08\u2009g\/mol=219.79\u2009g\\text{Mass of H\u2082SO\u2084} = 2.24 \\, \\text{mol} \\times 98.08 \\, \\text{g\/mol} = 219.79 \\, \\text{g}Mass\u00a0of\u00a0H\u2082SO\u2084=2.24mol\u00d798.08g\/mol=219.79g Thus, the mass of the solvent (water) is: Mass\u00a0of\u00a0water=1135\u2009g\u2212219.79\u2009g=915.21\u2009g\u22480.915\u2009kg\\text{Mass of water} = 1135 \\, \\text{g} – 219.79 \\, \\text{g} = 915.21 \\, \\text{g} \\approx 0.915 \\, \\text{kg}Mass\u00a0of\u00a0water=1135g\u2212219.79g=915.21g\u22480.915kg<\/li>\n\n\n\n- Calculate the moles of H\u2082SO\u2084 in 1 liter of solution<\/strong>:
From the molality, we already know that 2.24 moles of H\u2082SO\u2084 are in 1 kilogram of water. Since the mass of water in 1 liter of solution is 0.915 kg, the number of moles of H\u2082SO\u2084 in 1 liter is: Moles\u00a0of\u00a0H\u2082SO\u2084=2.24\u2009mol\/kg\u00d70.915\u2009kg=2.05\u2009mol\\text{Moles of H\u2082SO\u2084} = 2.24 \\, \\text{mol\/kg} \\times 0.915 \\, \\text{kg} = 2.05 \\, \\text{mol}Moles\u00a0of\u00a0H\u2082SO\u2084=2.24mol\/kg\u00d70.915kg=2.05mol<\/li>\n\n\n\n- Calculate the molar concentration (M)<\/strong>:
Molarity (M) is defined as the number of moles of solute per liter of solution. So, the molar concentration is: Molar\u00a0concentration=Moles\u00a0of\u00a0H\u2082SO\u2084Volume\u00a0of\u00a0solution\u00a0in\u00a0liters=2.05\u2009mol1\u2009L=2.05\u2009M\\text{Molar concentration} = \\frac{\\text{Moles of H\u2082SO\u2084}}{\\text{Volume of solution in liters}} = \\frac{2.05 \\, \\text{mol}}{1 \\, \\text{L}} = 2.05 \\, \\text{M}Molar\u00a0concentration=Volume\u00a0of\u00a0solution\u00a0in\u00a0litersMoles\u00a0of\u00a0H\u2082SO\u2084\u200b=1L2.05mol\u200b=2.05M<\/li>\n<\/ol>\n\n\n\nConclusion:<\/h3>\n\n\n\n
The molar concentration of the solution is 2.05 M<\/strong>.<\/p>\n\n\n\n
This means the solution contains 2.05 moles of H\u2082SO\u2084 per liter of solution.<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"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 find the molar concentration of the sulfuric acid (H\u2082SO\u2084) solution, we need to use the information given: Step-by-Step Solution: Conclusion: The […]<\/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-248284","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248284","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=248284"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248284\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248284"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248284"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248284"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Find the mass of the solution<\/strong>:
- Convert molality to moles of H\u2082SO\u2084<\/strong>: