{"id":187244,"date":"2025-02-03T19:01:26","date_gmt":"2025-02-03T19:01:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187244"},"modified":"2025-02-03T19:01:28","modified_gmt":"2025-02-03T19:01:28","slug":"find-the-amount-in-millimoles-of-the-indicated-species-in","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/03\/find-the-amount-in-millimoles-of-the-indicated-species-in\/","title":{"rendered":"Find the amount in millimoles of the indicated species in"},"content":{"rendered":"\n<p>Find the amount in millimoles of the indicated species in<\/p>\n\n\n\n<p>(a) 850 mg of P2O5.<\/p>\n\n\n\n<p>(b) 40.0 g of CO2.<\/p>\n\n\n\n<p>(c) 12.92 g of NaHCO3.<\/p>\n\n\n\n<p>(d) 57 mg of MgNH4PO4.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Find the number of millimoles of solute in<\/li>\n<\/ol>\n\n\n\n<p>(a) 2.00 L of 0.0555 M KMnO4.<\/p>\n\n\n\n<p>(b) 750 mL of 3.25 X 10-3 M KSCN.<\/p>\n\n\n\n<p>(c) 3.50 L of a solution that contains 3.33 ppm of CuSO4.<\/p>\n\n\n\n<p>(d) 250 mL of 0.414 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 determine the amount in millimoles (mmol) of a given species, we can use the following general approaches:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>For solid compounds<\/strong>: Convert the mass of the compound to moles by dividing by its molar mass, then multiply by 1,000 to convert to millimoles.<\/li>\n\n\n\n<li><strong>For solutions<\/strong>: Multiply the volume of the solution (in liters) by its molarity (in moles per liter), then multiply by 1,000 to convert to millimoles.<\/li>\n<\/ol>\n\n\n\n<p>Let&#8217;s apply these methods to each part of your question:<\/p>\n\n\n\n<p><strong>(a) 850 mg of P\u2082O\u2085<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Convert mass to grams<\/strong>: 50 mg = 0.850 g- <strong>Calculate molar mass of P\u2082O\u2085<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Phosphorus (P): 0.97 g\/mol &#8211; Oxygen (O): 6.00 g\/mol &#8211; Molar mass of P\u2082O\u2085 = 2 \u00d7 30.97) + (5 \u00d7 16.00) = 141.94 g\/mol- <strong>Calculate moles<\/strong>: .850 g \u00f7 141.94 g\/mol \u2248 0.00599 mol- <strong>Convert to millimoles<\/strong>: .00599 mol \u00d7 1,000 \u2248 5.99 mmol <strong>(b) 40.0 g of CO\u2082<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Calculate molar mass of CO\u2082<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 2.01 g\/mol &#8211; Oxygen (O): 6.00 g\/mol &#8211; Molar mass of CO\u2082 = 2.01 + (2 \u00d7 16.00) = 44.01 g\/mol- <strong>Calculate moles<\/strong>: 0.0 g \u00f7 44.01 g\/mol \u2248 0.909 mol- <strong>Convert to millimoles<\/strong>: .909 mol \u00d7 1,000 \u2248 909 mmol <strong>(c) 12.92 g of NaHCO\u2083<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Calculate molar mass of NaHCO\u2083<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Sodium (Na): 2.99 g\/mol &#8211; Hydrogen (H): .01 g\/mol &#8211; Carbon (C): 2.01 g\/mol &#8211; Oxygen (O): 6.00 g\/mol &#8211; Molar mass of NaHCO\u2083 = 2.99 + 1.01 + 12.01 + (3 \u00d7 16.00) = 84.01 g\/mol- <strong>Calculate moles<\/strong>: 2.92 g \u00f7 84.01 g\/mol \u2248 0.1538 mol- <strong>Convert to millimoles<\/strong>: .1538 mol \u00d7 1,000 \u2248 153.8 mmol <strong>(d) 57 mg of MgNH\u2084PO\u2084<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Convert mass to grams<\/strong>: 7 mg = 0.057 g- <strong>Calculate molar mass of MgNH\u2084PO\u2084<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Magnesium (Mg): 4.31 g\/mol &#8211; Nitrogen (N): 4.01 g\/mol &#8211; Hydrogen (H): .01 g\/mol &#8211; Phosphorus (P): 0.97 g\/mol &#8211; Oxygen (O): 6.00 g\/mol &#8211; Molar mass of MgNH\u2084PO\u2084 = 4.31 + 14.01 + (4 \u00d7 1.01) + 30.97 + (4 \u00d7 16.00) = 137.33 g\/mol- <strong>Calculate moles<\/strong>: .057 g \u00f7 137.33 g\/mol \u2248 0.000415 mol- <strong>Convert to millimoles<\/strong>: .000415 mol \u00d7 1,000 \u2248 0.415 mmol <strong>1. (a) 2.00 L of 0.0555 M KMnO\u2084<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Calculate moles<\/strong>: .00 L \u00d7 0.0555 mol\/L = 0.111 mol- <strong>Convert to millimoles<\/strong>: .111 mol \u00d7 1,000 = 111 mmol <strong>(b) 750 mL of 3.25 \u00d7 10\u207b\u00b3 M KSCN<\/strong><\/li>\n\n\n\n<li><strong>Convert volume to liters<\/strong>: 50 mL = 0.750 L- <strong>Calculate moles<\/strong>: .750 L \u00d7 3.25 \u00d7 10\u207b\u00b3 mol\/L = 0.0024375 mol- <strong>Convert to millimoles<\/strong>: .0024375 mol \u00d7 1,000 \u2248 2.44 mmol <strong>(c) 3.50 L of a solution that contains 3.33 ppm of CuSO\u2084<\/strong><\/li>\n\n\n\n<li><strong>Understand ppm<\/strong>: .33 ppm means 3.33 mg of CuSO\u2084 per liter of solution.- <strong>Total mass of CuSO\u2084<\/strong>: .33 mg\/L \u00d7 3.50 L = 11.655 mg- <strong>Convert mass to grams<\/strong>: 1.655 mg = 0.011655 g- <strong>Calculate molar mass of CuSO\u2084<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Copper (Cu): 3.55 g\/mol &#8211; Sulfur (S): 2.07 g\/mol &#8211; Oxygen (O): 6.00 g\/mol &#8211; Molar mass of CuSO\u2084 = 3.55 + 32.07 + (4 \u00d7 16.00) = 159.61 g\/mol- <strong>Calculate moles<\/strong>: .011655 g \u00f7 159<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Find the amount in millimoles of the indicated species in (a) 850 mg of P2O5. (b) 40.0 g of CO2. (c) 12.92 g of NaHCO3. (d) 57 mg of MgNH4PO4. (a) 2.00 L of 0.0555 M KMnO4. (b) 750 mL of 3.25 X 10-3 M KSCN. (c) 3.50 L of a solution that contains 3.33 [&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-187244","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187244","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=187244"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187244\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187244"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187244"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187244"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}