{"id":186718,"date":"2025-01-26T06:29:54","date_gmt":"2025-01-26T06:29:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186718"},"modified":"2025-01-26T06:29:58","modified_gmt":"2025-01-26T06:29:58","slug":"how-many-grams-of-barium-acetate-bach3coo2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/26\/how-many-grams-of-barium-acetate-bach3coo2\/","title":{"rendered":"How many grams of barium acetate, Ba(CH3COO)2"},"content":{"rendered":"\n<p>How many grams of barium acetate, Ba(CH3COO)2, must be dissolved to prepare 250. mL of a 0.218 M aqueous solution of the salt? <\/p>\n\n\n\n<p>How many grams of chromium(III) sulfate, Cr2(SO4)3, must be dissolved to prepare 100. mL of a 0.123 M aqueous solution of the salt?  <\/p>\n\n\n\n<p>How many mL of a 0.167 M aqueous solution of calcium bromide, CaBr2, must be taken to obtain 8.51 grams of the salt? mL<\/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\">1. Grams of Barium Acetate (Ba(CH\u2083COO)\u2082) for 250 mL of 0.218 M Solution<\/h3>\n\n\n\n<p><strong>Step 1: Calculate the moles of Ba(CH\u2083COO)\u2082 required.<\/strong><\/p>\n\n\n\n<p>The formula for molarity is:<\/p>\n\n\n\n<p>[<br>Molarity (M) = \\frac{moles\\ of\\ solute}{volume\\ of\\ solution\\ in\\ liters}<br>]<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molarity ( M = 0.218 \\, \\text{M} )<\/li>\n\n\n\n<li>Volume ( V = 250 \\, \\text{mL} = 0.250 \\, \\text{L} )<\/li>\n<\/ul>\n\n\n\n<p>Now, calculate the moles of Ba(CH\u2083COO)\u2082:<\/p>\n\n\n\n<p>[<br>\\text{moles} = M \\times V = 0.218 \\, \\text{mol\/L} \\times 0.250 \\, \\text{L} = 0.0545 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p><strong>Step 2: Calculate the mass of Ba(CH\u2083COO)\u2082.<\/strong><\/p>\n\n\n\n<p>The molar mass of Ba(CH\u2083COO)\u2082 can be calculated by adding the atomic masses of the elements:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ba: 137.33 g\/mol<\/li>\n\n\n\n<li>C: 12.01 g\/mol<\/li>\n\n\n\n<li>H: 1.008 g\/mol<\/li>\n\n\n\n<li>O: 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Molar mass of Ba(CH\u2083COO)\u2082:<\/p>\n\n\n\n<p>[<br>\\text{Molar mass} = 137.33 + 2 \\times (2 \\times 12.01 + 3 \\times 1.008 + 2 \\times 16.00) = 137.33 + 2 \\times 59.04 = 137.33 + 118.08 = 255.41 \\, \\text{g\/mol}<br>]<\/p>\n\n\n\n<p>Now, calculate the mass of Ba(CH\u2083COO)\u2082:<\/p>\n\n\n\n<p>[<br>\\text{mass} = \\text{moles} \\times \\text{molar mass} = 0.0545 \\, \\text{mol} \\times 255.41 \\, \\text{g\/mol} = 13.9 \\, \\text{g}<br>]<\/p>\n\n\n\n<p><strong>Answer:<\/strong><br>To prepare 250 mL of a 0.218 M solution of Ba(CH\u2083COO)\u2082, you need 13.9 grams of the salt.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. Grams of Chromium(III) Sulfate (Cr\u2082(SO\u2084)\u2083) for 100 mL of 0.123 M Solution<\/h3>\n\n\n\n<p><strong>Step 1: Calculate the moles of Cr\u2082(SO\u2084)\u2083 required.<\/strong><\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molarity ( M = 0.123 \\, \\text{M} )<\/li>\n\n\n\n<li>Volume ( V = 100 \\, \\text{mL} = 0.100 \\, \\text{L} )<\/li>\n<\/ul>\n\n\n\n<p>Calculate the moles of Cr\u2082(SO\u2084)\u2083:<\/p>\n\n\n\n<p>[<br>\\text{moles} = M \\times V = 0.123 \\, \\text{mol\/L} \\times 0.100 \\, \\text{L} = 0.0123 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p><strong>Step 2: Calculate the molar mass of Cr\u2082(SO\u2084)\u2083.<\/strong><\/p>\n\n\n\n<p>The molar mass of Cr\u2082(SO\u2084)\u2083 can be calculated by adding the atomic masses:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Cr: 52.00 g\/mol<\/li>\n\n\n\n<li>S: 32.07 g\/mol<\/li>\n\n\n\n<li>O: 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Molar mass of Cr\u2082(SO\u2084)\u2083:<\/p>\n\n\n\n<p>[<br>\\text{Molar mass} = 2 \\times 52.00 + 3 \\times (32.07 + 4 \\times 16.00) = 104.00 + 3 \\times 96.07 = 104.00 + 288.21 = 392.21 \\, \\text{g\/mol}<br>]<\/p>\n\n\n\n<p>Now, calculate the mass of Cr\u2082(SO\u2084)\u2083:<\/p>\n\n\n\n<p>[<br>\\text{mass} = \\text{moles} \\times \\text{molar mass} = 0.0123 \\, \\text{mol} \\times 392.21 \\, \\text{g\/mol} = 4.82 \\, \\text{g}<br>]<\/p>\n\n\n\n<p><strong>Answer:<\/strong><br>To prepare 100 mL of a 0.123 M solution of Cr\u2082(SO\u2084)\u2083, you need 4.82 grams of the salt.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3. Volume of 0.167 M CaBr\u2082 Solution to Obtain 8.51 Grams<\/h3>\n\n\n\n<p><strong>Step 1: Calculate the moles of CaBr\u2082 required.<\/strong><\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of CaBr\u2082 = 40.08 g\/mol (Ca) + 2 \u00d7 79.90 g\/mol (Br) = 40.08 + 159.80 = 199.88 g\/mol<\/li>\n\n\n\n<li>Mass of CaBr\u2082 = 8.51 g<\/li>\n<\/ul>\n\n\n\n<p>Now, calculate the moles of CaBr\u2082:<\/p>\n\n\n\n<p>[<br>\\text{moles} = \\frac{\\text{mass}}{\\text{molar mass}} = \\frac{8.51 \\, \\text{g}}{199.88 \\, \\text{g\/mol}} = 0.0426 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p><strong>Step 2: Calculate the volume of 0.167 M CaBr\u2082 solution required.<\/strong><\/p>\n\n\n\n<p>Use the molarity equation:<\/p>\n\n\n\n<p>[<br>M = \\frac{\\text{moles}}{\\text{volume}}<br>]<\/p>\n\n\n\n<p>Rearrange to solve for volume:<\/p>\n\n\n\n<p>[<br>\\text{volume} = \\frac{\\text{moles}}{M} = \\frac{0.0426 \\, \\text{mol}}{0.167 \\, \\text{mol\/L}} = 0.255 \\, \\text{L} = 255 \\, \\text{mL}<br>]<\/p>\n\n\n\n<p><strong>Answer:<\/strong><br>To obtain 8.51 grams of CaBr\u2082, you need to take 255 mL of a 0.167 M aqueous solution.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Summary of Answers:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>13.9 grams<\/strong> of Ba(CH\u2083COO)\u2082 for 250 mL of a 0.218 M solution.<\/li>\n\n\n\n<li><strong>4.82 grams<\/strong> of Cr\u2082(SO\u2084)\u2083 for 100 mL of a 0.123 M solution.<\/li>\n\n\n\n<li><strong>255 mL<\/strong> of a 0.167 M solution of CaBr\u2082 to obtain 8.51 grams of the salt.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>How many grams of barium acetate, Ba(CH3COO)2, must be dissolved to prepare 250. mL of a 0.218 M aqueous solution of the salt? How many grams of chromium(III) sulfate, Cr2(SO4)3, must be dissolved to prepare 100. mL of a 0.123 M aqueous solution of the salt? How many mL of a 0.167 M aqueous solution [&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-186718","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186718","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=186718"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186718\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186718"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186718"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186718"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}