{"id":232166,"date":"2025-06-11T18:52:35","date_gmt":"2025-06-11T18:52:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=232166"},"modified":"2025-06-11T18:52:39","modified_gmt":"2025-06-11T18:52:39","slug":"what-is-the-molecular-mass-of-k2co3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/what-is-the-molecular-mass-of-k2co3\/","title":{"rendered":"What is the molecular mass of K2CO3"},"content":{"rendered":"\n
What is the molecular mass of K2CO3 ? Determine the concentration in (a) molarity and (b) parts per million (ppm) of K2CO3 in a solution containing 3.60g of K2CO3 in a volume of 350mL of water<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Step 1: Molecular Mass of K\u2082CO\u2083<\/strong><\/h3>\n\n\n\n
The molecular formula for potassium carbonate is K\u2082CO\u2083<\/strong>, which consists of:<\/p>\n\n\n\n
Total molecular mass = 78.20 + 12.01 + 48.00 = 138.21 g\/mol<\/strong><\/p>\n\n\n\n\n\n\n\n
Step 2: Molarity of K\u2082CO\u2083 Solution<\/strong><\/h3>\n\n\n\n
Molarity (M) is defined as the number of moles of solute<\/strong> per liter of solution<\/strong>: Moles of K2CO3=3.60 g138.21 g\/mol=0.02605 mol\\text{Moles of } K_2CO_3 = \\frac{3.60\\ \\text{g}}{138.21\\ \\text{g\/mol}} = 0.02605\\ \\text{mol}Moles of K2\u200bCO3\u200b=138.21 g\/mol3.60 g\u200b=0.02605 mol<\/p>\n\n\n\n
Given the volume of solution is 350 mL = 0.350 L<\/strong>, the molarity is: M=0.02605 mol0.350 L=0.0744 mol\/LM = \\frac{0.02605\\ \\text{mol}}{0.350\\ \\text{L}} = 0.0744\\ \\text{mol\/L}M=0.350 L0.02605 mol\u200b=0.0744 mol\/L<\/p>\n\n\n\n\n\n\n\n
Step 3: Concentration in Parts Per Million (ppm)<\/strong><\/h3>\n\n\n\n
PPM is the mass of solute per million parts of solution (typically mg\/L for aqueous solutions): Mass of K2CO3=3.60 g=3600 mg\\text{Mass of } K_2CO_3 = 3.60\\ \\text{g} = 3600\\ \\text{mg}Mass of K2\u200bCO3\u200b=3.60 g=3600 mg Volume=350 mL=0.350 L\\text{Volume} = 350\\ \\text{mL} = 0.350\\ \\text{L}Volume=350 mL=0.350 L PPM=3600 mg0.350 L=10,\u2009\u2063285.71 ppm\\text{PPM} = \\frac{3600\\ \\text{mg}}{0.350\\ \\text{L}} = 10,\\!285.71\\ \\text{ppm}PPM=0.350 L3600 mg\u200b=10,285.71 ppm<\/p>\n\n\n\n
\n\n\n\n
Final Answers:<\/strong><\/h3>\n\n\n\n
\n
Molecular Mass of K\u2082CO\u2083<\/strong>: 138.21 g\/mol<\/li>\n\n\n\n
(a) Molarity<\/strong>: 0.0744 M<\/li>\n\n\n\n
(b) Concentration in ppm<\/strong>: 10,285.71 ppm<\/li>\n<\/ul>\n\n\n\n\n\n\n\n
Explanation<\/strong><\/h3>\n\n\n\n
Potassium carbonate (K\u2082CO\u2083) is an inorganic salt widely used in laboratory and industrial applications. Understanding its concentration in a solution is essential in chemical calculations involving stoichiometry, titration, and chemical reactions. The molecular mass, or molar mass, is determined by summing the atomic masses of all the atoms in the compound. Potassium contributes 39.10 g\/mol and there are two potassium atoms, along with one carbon atom and three oxygen atoms. This gives K\u2082CO\u2083 a total molar mass of 138.21 g\/mol.<\/p>\n\n\n\n
Molarity, a common concentration unit in chemistry, refers to the number of moles of solute per liter of solution. To find the number of moles in the 3.60 g of K\u2082CO\u2083, we divide by the molar mass. This gives us approximately 0.02605 moles. The solution volume is given as 350 mL, or 0.350 liters, so the molarity is calculated by dividing the number of moles by this volume, yielding 0.0744 M. This means that every liter of the solution would contain 0.0744 moles of potassium carbonate.<\/p>\n\n\n\n
To express concentration in parts per million (ppm), which is often used in environmental and water chemistry, we convert the mass of solute into milligrams. Since 3.60 g equals 3600 mg and the volume of the solution is 0.350 L, ppm is calculated as the milligrams of solute per liter of solution. Dividing 3600 mg by 0.350 L results in a concentration of approximately 10,285.71 ppm. This indicates a relatively high concentration of potassium carbonate in the solution. These calculations are fundamental for accurate chemical preparation and analysis.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
What is the molecular mass of K2CO3 ? Determine the concentration in (a) molarity and (b) parts per million (ppm) of K2CO3 in a solution containing 3.60g of K2CO3 in a volume of 350mL of water The Correct Answer and Explanation is: Step 1: Molecular Mass of K\u2082CO\u2083 The molecular formula for potassium carbonate is […]<\/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-232166","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232166","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=232166"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232166\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=232166"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=232166"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=232166"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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