Correct Answer:<\/strong><\/h3>\n\n\n\n4.77<\/strong><\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\nTo find the mass of potassium iodide (KI)<\/strong> in a solution, we follow a step-by-step process using the molarity formula<\/strong> and molar mass<\/strong>.<\/p>\n\n\n\n
\n\n\n\nStep 1: Understand What\u2019s Given<\/strong><\/h4>\n\n\n\n\n- Volume of the solution: 50.0 mL<\/li>\n\n\n\n
- Molarity (concentration): 0.575 M (mol\/L)<\/li>\n\n\n\n
- Molar mass of KI: 166 g\/mol<\/li>\n<\/ul>\n\n\n\n
We want to find the mass (in grams)<\/strong> of KI in this volume of solution.<\/p>\n\n\n\n
\n\n\n\nStep 2: Convert Volume to Liters<\/strong><\/h4>\n\n\n\nMolarity is defined in moles per liter<\/strong>, so we need to convert milliliters to liters: 50.0 mL=0.0500 L50.0\\ \\text{mL} = 0.0500\\ \\text{L}50.0 mL=0.0500 L<\/p>\n\n\n\n
\n\n\n\nStep 3: Use the Molarity Formula<\/strong><\/h4>\n\n\n\nMolarity (M) is defined as: M=moles of soluteliters of solutionM = \\frac{\\text{moles of solute}}{\\text{liters of solution}}M=liters of solutionmoles of solute\u200b<\/p>\n\n\n\n
Rearranged to find moles: moles of KI=M\u00d7volume in L=0.575 mol\/L\u00d70.0500 L=0.02875 mol\\text{moles of KI} = M \\times \\text{volume in L} = 0.575\\ \\text{mol\/L} \\times 0.0500\\ \\text{L} = 0.02875\\ \\text{mol}moles of KI=M\u00d7volume in L=0.575 mol\/L\u00d70.0500 L=0.02875 mol<\/p>\n\n\n\n
\n\n\n\nStep 4: Convert Moles to Grams<\/strong><\/h4>\n\n\n\nNow, use the molar mass of KI (166 g\/mol) to find the mass: mass=moles\u00d7molar mass=0.02875 mol\u00d7166 g\/mol=4.7725 g\\text{mass} = \\text{moles} \\times \\text{molar mass} = 0.02875\\ \\text{mol} \\times 166\\ \\text{g\/mol} = 4.7725\\ \\text{g}mass=moles\u00d7molar mass=0.02875 mol\u00d7166 g\/mol=4.7725 g<\/p>\n\n\n\n
\n\n\n\nStep 5: Round to 3 Significant Digits<\/strong><\/h4>\n\n\n\nSince the given values (50.0 mL and 0.575 M) have 3 significant digits, we must round the final answer accordingly: 4.77\\boxed{4.77}4.77\u200b<\/p>\n\n\n\n
\n\n\n\nThis is the mass of KI in 50.0 mL of a 0.575 M solution.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner9-105.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"What is the mass of potassium iodide (KI molar mass 166 g\/mol) in 50.0 mL of 0.575 M KI? Report your answer to three digits, for example, 0.567, 1.54, and do not include the unit. The Correct Answer and Explanation is: Correct Answer: 4.77 Explanation To find the mass of potassium iodide (KI) in a […]<\/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-227566","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227566","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=227566"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227566\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227566"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227566"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Explanation<\/strong><\/h3>\n\n\n\nTo find the mass of potassium iodide (KI)<\/strong> in a solution, we follow a step-by-step process using the molarity formula<\/strong> and molar mass<\/strong>.<\/p>\n\n\n\n
\n\n\n\nStep 1: Understand What\u2019s Given<\/strong><\/h4>\n\n\n\n\n- Volume of the solution: 50.0 mL<\/li>\n\n\n\n
- Molarity (concentration): 0.575 M (mol\/L)<\/li>\n\n\n\n
- Molar mass of KI: 166 g\/mol<\/li>\n<\/ul>\n\n\n\n
We want to find the mass (in grams)<\/strong> of KI in this volume of solution.<\/p>\n\n\n\n
\n\n\n\nStep 2: Convert Volume to Liters<\/strong><\/h4>\n\n\n\nMolarity is defined in moles per liter<\/strong>, so we need to convert milliliters to liters: 50.0 mL=0.0500 L50.0\\ \\text{mL} = 0.0500\\ \\text{L}50.0 mL=0.0500 L<\/p>\n\n\n\n
\n\n\n\nStep 3: Use the Molarity Formula<\/strong><\/h4>\n\n\n\nMolarity (M) is defined as: M=moles of soluteliters of solutionM = \\frac{\\text{moles of solute}}{\\text{liters of solution}}M=liters of solutionmoles of solute\u200b<\/p>\n\n\n\n
Rearranged to find moles: moles of KI=M\u00d7volume in L=0.575 mol\/L\u00d70.0500 L=0.02875 mol\\text{moles of KI} = M \\times \\text{volume in L} = 0.575\\ \\text{mol\/L} \\times 0.0500\\ \\text{L} = 0.02875\\ \\text{mol}moles of KI=M\u00d7volume in L=0.575 mol\/L\u00d70.0500 L=0.02875 mol<\/p>\n\n\n\n
\n\n\n\nStep 4: Convert Moles to Grams<\/strong><\/h4>\n\n\n\nNow, use the molar mass of KI (166 g\/mol) to find the mass: mass=moles\u00d7molar mass=0.02875 mol\u00d7166 g\/mol=4.7725 g\\text{mass} = \\text{moles} \\times \\text{molar mass} = 0.02875\\ \\text{mol} \\times 166\\ \\text{g\/mol} = 4.7725\\ \\text{g}mass=moles\u00d7molar mass=0.02875 mol\u00d7166 g\/mol=4.7725 g<\/p>\n\n\n\n
\n\n\n\nStep 5: Round to 3 Significant Digits<\/strong><\/h4>\n\n\n\nSince the given values (50.0 mL and 0.575 M) have 3 significant digits, we must round the final answer accordingly: 4.77\\boxed{4.77}4.77\u200b<\/p>\n\n\n\n
\n\n\n\nThis is the mass of KI in 50.0 mL of a 0.575 M solution.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"What is the mass of potassium iodide (KI molar mass 166 g\/mol) in 50.0 mL of 0.575 M KI? Report your answer to three digits, for example, 0.567, 1.54, and do not include the unit. The Correct Answer and Explanation is: Correct Answer: 4.77 Explanation To find the mass of potassium iodide (KI) in a […]<\/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-227566","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227566","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=227566"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227566\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227566"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227566"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Step 1: Understand What\u2019s Given<\/strong><\/h4>\n\n\n\n\n- Volume of the solution: 50.0 mL<\/li>\n\n\n\n
- Molarity (concentration): 0.575 M (mol\/L)<\/li>\n\n\n\n
- Molar mass of KI: 166 g\/mol<\/li>\n<\/ul>\n\n\n\n
We want to find the mass (in grams)<\/strong> of KI in this volume of solution.<\/p>\n\n\n\n
\n\n\n\nStep 2: Convert Volume to Liters<\/strong><\/h4>\n\n\n\nMolarity is defined in moles per liter<\/strong>, so we need to convert milliliters to liters: 50.0 mL=0.0500 L50.0\\ \\text{mL} = 0.0500\\ \\text{L}50.0 mL=0.0500 L<\/p>\n\n\n\n
\n\n\n\nStep 3: Use the Molarity Formula<\/strong><\/h4>\n\n\n\nMolarity (M) is defined as: M=moles of soluteliters of solutionM = \\frac{\\text{moles of solute}}{\\text{liters of solution}}M=liters of solutionmoles of solute\u200b<\/p>\n\n\n\n
Rearranged to find moles: moles of KI=M\u00d7volume in L=0.575 mol\/L\u00d70.0500 L=0.02875 mol\\text{moles of KI} = M \\times \\text{volume in L} = 0.575\\ \\text{mol\/L} \\times 0.0500\\ \\text{L} = 0.02875\\ \\text{mol}moles of KI=M\u00d7volume in L=0.575 mol\/L\u00d70.0500 L=0.02875 mol<\/p>\n\n\n\n
\n\n\n\nStep 4: Convert Moles to Grams<\/strong><\/h4>\n\n\n\nNow, use the molar mass of KI (166 g\/mol) to find the mass: mass=moles\u00d7molar mass=0.02875 mol\u00d7166 g\/mol=4.7725 g\\text{mass} = \\text{moles} \\times \\text{molar mass} = 0.02875\\ \\text{mol} \\times 166\\ \\text{g\/mol} = 4.7725\\ \\text{g}mass=moles\u00d7molar mass=0.02875 mol\u00d7166 g\/mol=4.7725 g<\/p>\n\n\n\n
\n\n\n\nStep 5: Round to 3 Significant Digits<\/strong><\/h4>\n\n\n\nSince the given values (50.0 mL and 0.575 M) have 3 significant digits, we must round the final answer accordingly: 4.77\\boxed{4.77}4.77\u200b<\/p>\n\n\n\n
\n\n\n\nThis is the mass of KI in 50.0 mL of a 0.575 M solution.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"What is the mass of potassium iodide (KI molar mass 166 g\/mol) in 50.0 mL of 0.575 M KI? Report your answer to three digits, for example, 0.567, 1.54, and do not include the unit. The Correct Answer and Explanation is: Correct Answer: 4.77 Explanation To find the mass of potassium iodide (KI) in a […]<\/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-227566","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227566","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=227566"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227566\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227566"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227566"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
We want to find the mass (in grams)<\/strong> of KI in this volume of solution.<\/p>\n\n\n\n Molarity is defined in moles per liter<\/strong>, so we need to convert milliliters to liters: 50.0 mL=0.0500 L50.0\\ \\text{mL} = 0.0500\\ \\text{L}50.0 mL=0.0500 L<\/p>\n\n\n\n Molarity (M) is defined as: M=moles of soluteliters of solutionM = \\frac{\\text{moles of solute}}{\\text{liters of solution}}M=liters of solutionmoles of solute\u200b<\/p>\n\n\n\n Rearranged to find moles: moles of KI=M\u00d7volume in L=0.575 mol\/L\u00d70.0500 L=0.02875 mol\\text{moles of KI} = M \\times \\text{volume in L} = 0.575\\ \\text{mol\/L} \\times 0.0500\\ \\text{L} = 0.02875\\ \\text{mol}moles of KI=M\u00d7volume in L=0.575 mol\/L\u00d70.0500 L=0.02875 mol<\/p>\n\n\n\n Now, use the molar mass of KI (166 g\/mol) to find the mass: mass=moles\u00d7molar mass=0.02875 mol\u00d7166 g\/mol=4.7725 g\\text{mass} = \\text{moles} \\times \\text{molar mass} = 0.02875\\ \\text{mol} \\times 166\\ \\text{g\/mol} = 4.7725\\ \\text{g}mass=moles\u00d7molar mass=0.02875 mol\u00d7166 g\/mol=4.7725 g<\/p>\n\n\n\n Since the given values (50.0 mL and 0.575 M) have 3 significant digits, we must round the final answer accordingly: 4.77\\boxed{4.77}4.77\u200b<\/p>\n\n\n\n This is the mass of KI in 50.0 mL of a 0.575 M solution.<\/p>\n\n\n\n What is the mass of potassium iodide (KI molar mass 166 g\/mol) in 50.0 mL of 0.575 M KI? Report your answer to three digits, for example, 0.567, 1.54, and do not include the unit. The Correct Answer and Explanation is: Correct Answer: 4.77 Explanation To find the mass of potassium iodide (KI) in a […]<\/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-227566","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227566","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=227566"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227566\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227566"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227566"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 2: Convert Volume to Liters<\/strong><\/h4>\n\n\n\n
\n\n\n\nStep 3: Use the Molarity Formula<\/strong><\/h4>\n\n\n\n
\n\n\n\nStep 4: Convert Moles to Grams<\/strong><\/h4>\n\n\n\n
\n\n\n\nStep 5: Round to 3 Significant Digits<\/strong><\/h4>\n\n\n\n
\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"