{"id":209174,"date":"2025-04-28T14:01:40","date_gmt":"2025-04-28T14:01:40","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=209174"},"modified":"2025-04-28T14:01:43","modified_gmt":"2025-04-28T14:01:43","slug":"how-many-millimoles-of-solute-are-contained-in","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/28\/how-many-millimoles-of-solute-are-contained-in\/","title":{"rendered":"How many millimoles of solute are contained in"},"content":{"rendered":"\n<p>How many millimoles of solute are contained in<\/p>\n\n\n\n<p>(a) 2.00 L of 2.76 X 10-3 M KMnO4?<\/p>\n\n\n\n<p>(b) 250.0 mL of 0.0423 M KSCN?<\/p>\n\n\n\n<p>(c) 500.0 mL of a solution containing 2.97 ppm CuSO4?<\/p>\n\n\n\n<p>(d) 2.50 L of 0.352 M KCl?<\/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 calculate the number of millimoles (mmol) of solute, we use the following formula:<\/p>\n\n\n\n<p>[<br>\\text{Millimoles of solute} = \\text{Concentration (in mol\/L)} \\times \\text{Volume (in L)} \\times 1000<br>]<\/p>\n\n\n\n<p>The factor of 1000 is used to convert moles to millimoles (since 1 mol = 1000 mmol).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(a) 2.00 L of 2.76 \u00d7 10\u207b\u00b3 M KMnO\u2084<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Concentration (M) = 2.76 \u00d7 10\u207b\u00b3 M<\/li>\n\n\n\n<li>Volume (L) = 2.00 L<\/li>\n<\/ul>\n\n\n\n<p>First, we apply the formula:<br>[<br>\\text{Millimoles of KMnO\u2084} = 2.76 \\times 10^{-3} \\, \\text{M} \\times 2.00 \\, \\text{L} \\times 1000 = 5.52 \\, \\text{mmol}<br>]<\/p>\n\n\n\n<p>So, the number of millimoles of KMnO\u2084 in 2.00 L of solution is <strong>5.52 mmol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(b) 250.0 mL of 0.0423 M KSCN<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Concentration (M) = 0.0423 M<\/li>\n\n\n\n<li>Volume (mL) = 250.0 mL = 0.2500 L (since 1000 mL = 1 L)<\/li>\n<\/ul>\n\n\n\n<p>Using the formula:<br>[<br>\\text{Millimoles of KSCN} = 0.0423 \\, \\text{M} \\times 0.2500 \\, \\text{L} \\times 1000 = 10.575 \\, \\text{mmol}<br>]<\/p>\n\n\n\n<p>So, the number of millimoles of KSCN in 250.0 mL of solution is <strong>10.575 mmol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(c) 500.0 mL of a solution containing 2.97 ppm CuSO\u2084<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Concentration in ppm (parts per million) = 2.97 ppm<\/li>\n\n\n\n<li>Volume (mL) = 500.0 mL = 0.5000 L<\/li>\n\n\n\n<li>1 ppm is equivalent to 1 mg\/L, so the concentration is 2.97 mg\/L.<\/li>\n<\/ul>\n\n\n\n<p>To convert the concentration from milligrams to moles, we need to use the molar mass of CuSO\u2084. The molar mass of CuSO\u2084 is:<br>[<br>\\text{Molar mass of CuSO\u2084} = 63.55 (\\text{Cu}) + 32.07 (\\text{S}) + 4 \\times 16.00 (\\text{O}) = 159.62 \\, \\text{g\/mol}<br>]<\/p>\n\n\n\n<p>Now, convert the concentration to moles per liter:<br>[<br>\\text{Concentration (mol\/L)} = \\frac{2.97 \\, \\text{mg\/L}}{159.62 \\, \\text{g\/mol}} = \\frac{2.97 \\times 10^{-3} \\, \\text{g\/L}}{159.62 \\, \\text{g\/mol}} = 1.86 \\times 10^{-5} \\, \\text{mol\/L}<br>]<\/p>\n\n\n\n<p>Now, calculate the millimoles:<br>[<br>\\text{Millimoles of CuSO\u2084} = 1.86 \\times 10^{-5} \\, \\text{mol\/L} \\times 0.5000 \\, \\text{L} \\times 1000 = 0.0093 \\, \\text{mmol}<br>]<\/p>\n\n\n\n<p>So, the number of millimoles of CuSO\u2084 in 500.0 mL of solution is <strong>0.0093 mmol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(d) 2.50 L of 0.352 M KCl<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Concentration (M) = 0.352 M<\/li>\n\n\n\n<li>Volume (L) = 2.50 L<\/li>\n<\/ul>\n\n\n\n<p>Using the formula:<br>[<br>\\text{Millimoles of KCl} = 0.352 \\, \\text{M} \\times 2.50 \\, \\text{L} \\times 1000 = 880 \\, \\text{mmol}<br>]<\/p>\n\n\n\n<p>So, the number of millimoles of KCl in 2.50 L of solution is <strong>880 mmol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(a) <strong>5.52 mmol<\/strong> of KMnO\u2084<\/li>\n\n\n\n<li>(b) <strong>10.575 mmol<\/strong> of KSCN<\/li>\n\n\n\n<li>(c) <strong>0.0093 mmol<\/strong> of CuSO\u2084<\/li>\n\n\n\n<li>(d) <strong>880 mmol<\/strong> of KCl<\/li>\n<\/ul>\n\n\n\n<p>These calculations demonstrate how to convert molarity and volume into the number of millimoles of solute present in a solution.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How many millimoles of solute are contained in (a) 2.00 L of 2.76 X 10-3 M KMnO4? (b) 250.0 mL of 0.0423 M KSCN? (c) 500.0 mL of a solution containing 2.97 ppm CuSO4? (d) 2.50 L of 0.352 M KCl? The correct answer and explanation is : To calculate the number of millimoles (mmol) [&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-209174","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209174","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=209174"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209174\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209174"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209174"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209174"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}