{"id":251853,"date":"2025-07-10T20:17:47","date_gmt":"2025-07-10T20:17:47","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=251853"},"modified":"2025-07-10T20:17:51","modified_gmt":"2025-07-10T20:17:51","slug":"determine-the-molarity-for-0-2074-g-of-calcium-hydroxide-caoh2-in-40-00-ml-of-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/10\/determine-the-molarity-for-0-2074-g-of-calcium-hydroxide-caoh2-in-40-00-ml-of-solution\/","title":{"rendered":"Determine the molarity for 0.2074 g of calcium hydroxide, Ca(OH)2, in 40.00 mL of solution."},"content":{"rendered":"\n<p>Determine the molarity for 0.2074 g of calcium hydroxide, Ca(OH)2, in 40.00 mL of solution. If you know the molar mass of H=1, O=16, and Ca=40 g\/mol.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the molarity of calcium hydroxide (Ca(OH)2) in a solution, follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the molar mass of Ca(OH)2.<\/h3>\n\n\n\n<p>The formula for calcium hydroxide is Ca(OH)2. We can calculate its molar mass by adding the atomic masses of all the atoms in the compound:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of Ca = 40 g\/mol<\/li>\n\n\n\n<li>Molar mass of O = 16 g\/mol<\/li>\n\n\n\n<li>Molar mass of H = 1 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>For one molecule of Ca(OH)2:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ca = 1 \u00d7 40 g\/mol = 40 g\/mol<\/li>\n\n\n\n<li>O = 2 \u00d7 16 g\/mol = 32 g\/mol<\/li>\n\n\n\n<li>H = 2 \u00d7 1 g\/mol = 2 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>The total molar mass of Ca(OH)2 is:<br>40+32+2=74\u2009g\/mol40 + 32 + 2 = 74 \\, \\text{g\/mol}40+32+2=74g\/mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Convert the mass of Ca(OH)2 to moles.<\/h3>\n\n\n\n<p>The given mass of Ca(OH)2 is 0.2074 g. To find the number of moles of Ca(OH)2, use the formula: moles=massmolar&nbsp;mass\\text{moles} = \\frac{\\text{mass}}{\\text{molar mass}}moles=molar&nbsp;massmass\u200b<\/p>\n\n\n\n<p>Substituting the known values: moles&nbsp;of&nbsp;Ca(OH)2=0.2074\u2009g74\u2009g\/mol=0.00280\u2009mol\\text{moles of Ca(OH)2} = \\frac{0.2074 \\, \\text{g}}{74 \\, \\text{g\/mol}} = 0.00280 \\, \\text{mol}moles&nbsp;of&nbsp;Ca(OH)2=74g\/mol0.2074g\u200b=0.00280mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Convert the volume of the solution to liters.<\/h3>\n\n\n\n<p>The volume of the solution is given as 40.00 mL. Since molarity is defined in terms of liters of solution, convert milliliters to liters: 40.00\u2009mL=40.00\u2009mL\u00d71\u2009L1000\u2009mL=0.04000\u2009L40.00 \\, \\text{mL} = 40.00 \\, \\text{mL} \\times \\frac{1 \\, \\text{L}}{1000 \\, \\text{mL}} = 0.04000 \\, \\text{L}40.00mL=40.00mL\u00d71000mL1L\u200b=0.04000L<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate the molarity.<\/h3>\n\n\n\n<p>Molarity (M) is defined as the number of moles of solute per liter of solution. Using the formula: M=moles&nbsp;of&nbsp;solutevolume&nbsp;of&nbsp;solution&nbsp;in&nbsp;litersM = \\frac{\\text{moles of solute}}{\\text{volume of solution in liters}}M=volume&nbsp;of&nbsp;solution&nbsp;in&nbsp;litersmoles&nbsp;of&nbsp;solute\u200b<\/p>\n\n\n\n<p>Substitute the values: M=0.00280\u2009mol0.04000\u2009L=0.0700\u2009MM = \\frac{0.00280 \\, \\text{mol}}{0.04000 \\, \\text{L}} = 0.0700 \\, \\text{M}M=0.04000L0.00280mol\u200b=0.0700M<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The molarity of the Ca(OH)2 solution is <strong>0.0700 M<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The molarity tells us the concentration of the Ca(OH)2 solution in moles per liter. By calculating the moles of Ca(OH)2 from its mass and then dividing by the volume of the solution, we determine that the concentration is 0.0700 M.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-65.jpeg\" alt=\"\" class=\"wp-image-251854\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the molarity for 0.2074 g of calcium hydroxide, Ca(OH)2, in 40.00 mL of solution. If you know the molar mass of H=1, O=16, and Ca=40 g\/mol. The Correct Answer and Explanation is: To calculate the molarity of calcium hydroxide (Ca(OH)2) in a solution, follow these steps: Step 1: Find the molar mass of Ca(OH)2. [&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-251853","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251853","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=251853"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251853\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=251853"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=251853"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=251853"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}