{"id":182152,"date":"2025-01-13T12:53:07","date_gmt":"2025-01-13T12:53:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182152"},"modified":"2025-01-13T12:53:09","modified_gmt":"2025-01-13T12:53:09","slug":"solute-mass-of-solute-moles-of-solute-volume-of-solution-molarity-of-solution-mgso4-0-598-g-_-20-0-ml-naoh-110-0-ml-1-75-m-ch3oh-12-0-g-_-0-550-m-a","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/solute-mass-of-solute-moles-of-solute-volume-of-solution-molarity-of-solution-mgso4-0-598-g-_-20-0-ml-naoh-110-0-ml-1-75-m-ch3oh-12-0-g-_-0-550-m-a\/","title":{"rendered":"Solute Mass of solute Moles of solute Volume of solution Molarity of solution MgSO4 0.598 g _ 20.0 mL NaOH 110.0 mL 1.75 M CH3OH 12.0 g _ 0.550 M A)"},"content":{"rendered":"\n<p>Complete the table: Solute Mass of solute Moles of solute Volume of solution Molarity of solution MgSO4 0.598 g _____ 20.0 mL _____ NaOH _____ _____ 110.0 mL 1.75 M CH3OH 12.0 g _____ _____ 0.550 M A) Complete the row for MgSO4 B) Complete the row for NaOH C) Complete the row for CH3OH<\/p>\n\n\n\n<p>Complete the table:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Solute<\/td><td>Mass of solute<\/td><td>Moles of solute<\/td><td>Volume of solution<\/td><td>Molarity of solution<\/td><\/tr><tr><td><strong>MgSO<sub>4<\/sub><\/strong><\/td><td>0.598 g<\/td><td>_____<\/td><td>20.0 mL<\/td><td>_____<\/td><\/tr><tr><td><strong>NaOH<\/strong><\/td><td>_____<\/td><td>_____<\/td><td>110.0 mL<\/td><td>1.75 M<\/td><\/tr><tr><td><strong>CH<sub>3<\/sub>OH<\/strong><\/td><td>12.0 g<\/td><td>_____<\/td><td>_____<\/td><td>0.550 M<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>A) Complete the row for&nbsp;<strong>MgSO<sub>4<\/sub><\/strong><\/p>\n\n\n\n<p>B) Complete the row for&nbsp;<strong>NaOH<\/strong><\/p>\n\n\n\n<p>C) Complete the row for&nbsp;<strong>CH<sub>3<\/sub>OH<\/strong><\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To complete the table, we need to calculate the missing values for each solute using the relationships between mass, moles, volume, and molarity.<\/p>\n\n\n\n<p><strong>A) MgSO\u2084<\/strong><\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of solute: 0.598 g<\/li>\n\n\n\n<li>Volume of solution: 20.0 mL (0.0200 L)<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Moles of MgSO\u2084<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of MgSO\u2084 = 120.37 g\/mol<\/li>\n\n\n\n<li>Moles = Mass \/ Molar mass<\/li>\n\n\n\n<li>Moles = 0.598 g \/ 120.37 g\/mol \u2248 0.00497 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Molarity of solution<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molarity (M) = Moles \/ Volume (L)<\/li>\n\n\n\n<li>Molarity = 0.00497 mol \/ 0.0200 L \u2248 0.2485 M<\/li>\n<\/ul>\n\n\n\n<p><strong>B) NaOH<\/strong><\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molarity of solution: 1.75 M<\/li>\n\n\n\n<li>Volume of solution: 110.0 mL (0.1100 L)<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Moles of NaOH<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles = Molarity \u00d7 Volume (L)<\/li>\n\n\n\n<li>Moles = 1.75 M \u00d7 0.1100 L = 0.1925 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Mass of NaOH<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of NaOH = 40.00 g\/mol<\/li>\n\n\n\n<li>Mass = Moles \u00d7 Molar mass<\/li>\n\n\n\n<li>Mass = 0.1925 mol \u00d7 40.00 g\/mol = 7.70 g<\/li>\n<\/ul>\n\n\n\n<p><strong>C) CH\u2083OH<\/strong><\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of solute: 12.0 g<\/li>\n\n\n\n<li>Molarity of solution: 0.550 M<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Moles of CH\u2083OH<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of CH\u2083OH = 32.04 g\/mol<\/li>\n\n\n\n<li>Moles = Mass \/ Molar mass<\/li>\n\n\n\n<li>Moles = 12.0 g \/ 32.04 g\/mol \u2248 0.3745 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Volume of solution<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume (L) = Moles \/ Molarity<\/li>\n\n\n\n<li>Volume = 0.3745 mol \/ 0.550 M \u2248 0.6809 L<\/li>\n\n\n\n<li>Volume = 0.6809 L \u00d7 1000 mL\/L \u2248 681 mL<\/li>\n<\/ul>\n\n\n\n<p><strong>Summary Table:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Solute<\/th><th>Mass of solute<\/th><th>Moles of solute<\/th><th>Volume of solution<\/th><th>Molarity of solution<\/th><\/tr><\/thead><tbody><tr><td>MgSO\u2084<\/td><td>0.598 g<\/td><td>0.00497 mol<\/td><td>20.0 mL (0.0200 L)<\/td><td>0.2485 M<\/td><\/tr><tr><td>NaOH<\/td><td>7.70 g<\/td><td>0.1925 mol<\/td><td>110.0 mL (0.1100 L)<\/td><td>1.75 M<\/td><\/tr><tr><td>CH\u2083OH<\/td><td>12.0 g<\/td><td>0.3745 mol<\/td><td>681 mL (0.681 L)<\/td><td>0.550 M<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Moles of Solute<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Calculated by dividing the mass of the solute by its molar mass.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Molarity of Solution<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Determined by dividing the moles of solute by the volume of the solution in liters.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Volume of Solution<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Found by dividing the moles of solute by the molarity of the solution.<\/li>\n<\/ul>\n\n\n\n<p>These calculations are essential for understanding solution concentrations and preparing solutions with desired molarities.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Complete the table: Solute Mass of solute Moles of solute Volume of solution Molarity of solution MgSO4 0.598 g _____ 20.0 mL _____ NaOH _____ _____ 110.0 mL 1.75 M CH3OH 12.0 g _____ _____ 0.550 M A) Complete the row for MgSO4 B) Complete the row for NaOH C) Complete the row for CH3OH [&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-182152","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182152","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=182152"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182152\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182152"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182152"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182152"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}