{"id":204814,"date":"2025-03-22T15:23:39","date_gmt":"2025-03-22T15:23:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=204814"},"modified":"2025-03-22T15:23:42","modified_gmt":"2025-03-22T15:23:42","slug":"calculate-the-concentration-of-a-standard-solution-of-hydrated-oxalic-acid-h2c2o4%c2%b72h2o-prepared-by-dissolving-25-21g-of-hydrated-oxalic-acid-in-250-0ml-of-deionised-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/22\/calculate-the-concentration-of-a-standard-solution-of-hydrated-oxalic-acid-h2c2o4%c2%b72h2o-prepared-by-dissolving-25-21g-of-hydrated-oxalic-acid-in-250-0ml-of-deionised-water\/","title":{"rendered":"Calculate the concentration of a standard solution of hydrated oxalic acid (H2C2O4\u00b72H2O) prepared by dissolving 25.21g of hydrated oxalic acid in 250.0mL of deionised water"},"content":{"rendered":"\n<p>&nbsp;Calculate the concentration of a standard solution of hydrated oxalic acid (H2C2O4\u00b72H2O) prepared by dissolving 25.21g of hydrated oxalic acid in 250.0mL of deionised water.<\/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<h3 class=\"wp-block-heading\"><strong>Calculation of the Concentration of Hydrated Oxalic Acid Solution<\/strong><\/h3>\n\n\n\n<p>To calculate the concentration of the standard solution of hydrated oxalic acid ((\\text{H}_2\\text{C}_2\\text{O}_4 \\cdot 2\\text{H}_2\\text{O})), we follow these steps:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 1: Determine the Molar Mass of Hydrated Oxalic Acid<\/strong><\/h4>\n\n\n\n<p>The molar masses of the constituent elements are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hydrogen (H) = (1.008 \\times 4 = 4.032) g\/mol<\/li>\n\n\n\n<li>Carbon (C) = (12.01 \\times 2 = 24.02) g\/mol<\/li>\n\n\n\n<li>Oxygen (O) = (16.00 \\times 6 = 96.00) g\/mol<\/li>\n\n\n\n<li>Water of hydration ((2\\text{H}_2\\text{O})) = (18.02 \\times 2 = 36.04) g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Total molar mass of (\\text{H}_2\\text{C}_2\\text{O}_4 \\cdot 2\\text{H}_2\\text{O}):<br>[<br>4.032 + 24.02 + 96.00 + 36.04 = 126.09 \\text{ g\/mol}<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 2: Calculate the Number of Moles of Hydrated Oxalic Acid<\/strong><\/h4>\n\n\n\n<p>Using the given mass of 25.21 g:<br>[<br>\\text{Moles} = \\frac{\\text{Mass}}{\\text{Molar Mass}} = \\frac{25.21}{126.09} = 0.2 \\text{ mol}<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 3: Calculate the Molar Concentration<\/strong><\/h4>\n\n\n\n<p>Concentration ((C)) is given by:<\/p>\n\n\n\n<p>[<br>C = \\frac{\\text{Moles of solute}}{\\text{Volume of solution in liters}}<br>]<\/p>\n\n\n\n<p>Since the solution volume is 250.0 mL = 0.2500 L,<\/p>\n\n\n\n<p>[<br>C = \\frac{0.2}{0.2500} = 0.800 \\text{ mol\/L}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The concentration of a solution refers to the amount of solute (in moles) dissolved in a given volume of solvent (in liters). In this case, the solute is hydrated oxalic acid, and the solvent is deionized water.<\/p>\n\n\n\n<p>We determined the concentration by:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Finding the molar mass<\/strong> of hydrated oxalic acid to understand how many grams constitute one mole.<\/li>\n\n\n\n<li><strong>Calculating the number of moles<\/strong> in the given mass of 25.21 g.<\/li>\n\n\n\n<li><strong>Dividing the moles by the solution volume<\/strong> to obtain the molarity.<\/li>\n<\/ol>\n\n\n\n<p>The result, <strong>0.800 mol\/L (M)<\/strong>, means there are <strong>0.800 moles of hydrated oxalic acid per liter<\/strong> of solution. This standard solution can now be used in titrations or other quantitative chemical analyses.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>&nbsp;Calculate the concentration of a standard solution of hydrated oxalic acid (H2C2O4\u00b72H2O) prepared by dissolving 25.21g of hydrated oxalic acid in 250.0mL of deionised water. The correct answer and explanation is : Calculation of the Concentration of Hydrated Oxalic Acid Solution To calculate the concentration of the standard solution of hydrated oxalic acid ((\\text{H}_2\\text{C}_2\\text{O}_4 \\cdot [&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-204814","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204814","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=204814"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204814\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204814"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204814"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204814"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}