{"id":254905,"date":"2025-07-12T17:31:25","date_gmt":"2025-07-12T17:31:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=254905"},"modified":"2025-07-12T17:31:34","modified_gmt":"2025-07-12T17:31:34","slug":"calculate-the-average-molar-mass-of","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/12\/calculate-the-average-molar-mass-of\/","title":{"rendered":"Calculate the \u201caverage\u201d molar mass of"},"content":{"rendered":"\n<p>Calculate the \u201caverage\u201d molar mass of <\/p>\n\n\n\n<p>(a) chlorobenzene (C6H5Cl) and <\/p>\n\n\n\n<p>(b) bromobenzene (C6H5Br<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the average molar mass of chlorobenzene (C\u2086H\u2085Cl) and bromobenzene (C\u2086H\u2085Br), we need to first determine the molar mass of each compound by summing the atomic masses of all the elements in their molecular formulas.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(a) <strong>Chlorobenzene (C\u2086H\u2085Cl)<\/strong><\/h3>\n\n\n\n<p>The molecular formula for chlorobenzene is C\u2086H\u2085Cl, which consists of 6 carbon (C) atoms, 5 hydrogen (H) atoms, and 1 chlorine (Cl) atom.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong>: The atomic mass of carbon is 12.01 g\/mol.\n<ul class=\"wp-block-list\">\n<li>For 6 carbon atoms: 6\u00d712.01=72.06\u2009g\/mol6 \\times 12.01 = 72.06 \\, \\text{g\/mol}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong>: The atomic mass of hydrogen is 1.008 g\/mol.\n<ul class=\"wp-block-list\">\n<li>For 5 hydrogen atoms: 5\u00d71.008=5.04\u2009g\/mol5 \\times 1.008 = 5.04 \\, \\text{g\/mol}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Chlorine (Cl)<\/strong>: The atomic mass of chlorine is 35.45 g\/mol.\n<ul class=\"wp-block-list\">\n<li>For 1 chlorine atom: 1\u00d735.45=35.45\u2009g\/mol1 \\times 35.45 = 35.45 \\, \\text{g\/mol}<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Now, add up the individual contributions: Molar&nbsp;mass&nbsp;of&nbsp;chlorobenzene=72.06+5.04+35.45=112.55\u2009g\/mol\\text{Molar mass of chlorobenzene} = 72.06 + 5.04 + 35.45 = 112.55 \\, \\text{g\/mol}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(b) <strong>Bromobenzene (C\u2086H\u2085Br)<\/strong><\/h3>\n\n\n\n<p>The molecular formula for bromobenzene is C\u2086H\u2085Br, consisting of 6 carbon (C) atoms, 5 hydrogen (H) atoms, and 1 bromine (Br) atom.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong>: The atomic mass of carbon is 12.01 g\/mol.\n<ul class=\"wp-block-list\">\n<li>For 6 carbon atoms: 6\u00d712.01=72.06\u2009g\/mol6 \\times 12.01 = 72.06 \\, \\text{g\/mol}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong>: The atomic mass of hydrogen is 1.008 g\/mol.\n<ul class=\"wp-block-list\">\n<li>For 5 hydrogen atoms: 5\u00d71.008=5.04\u2009g\/mol5 \\times 1.008 = 5.04 \\, \\text{g\/mol}<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Bromine (Br)<\/strong>: The atomic mass of bromine is 79.90 g\/mol.\n<ul class=\"wp-block-list\">\n<li>For 1 bromine atom: 1\u00d779.90=79.90\u2009g\/mol1 \\times 79.90 = 79.90 \\, \\text{g\/mol}<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Now, add up the individual contributions: Molar&nbsp;mass&nbsp;of&nbsp;bromobenzene=72.06+5.04+79.90=157.00\u2009g\/mol\\text{Molar mass of bromobenzene} = 72.06 + 5.04 + 79.90 = 157.00 \\, \\text{g\/mol}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Chlorobenzene (C\u2086H\u2085Cl)<\/strong> has a molar mass of <strong>112.55 g\/mol<\/strong>.<\/li>\n\n\n\n<li><strong>Bromobenzene (C\u2086H\u2085Br)<\/strong> has a molar mass of <strong>157.00 g\/mol<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>This shows that bromobenzene has a significantly higher molar mass compared to chlorobenzene due to the larger atomic mass of bromine compared to chlorine.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the \u201caverage\u201d molar mass of (a) chlorobenzene (C6H5Cl) and (b) bromobenzene (C6H5Br The correct answer and explanation is: To calculate the average molar mass of chlorobenzene (C\u2086H\u2085Cl) and bromobenzene (C\u2086H\u2085Br), we need to first determine the molar mass of each compound by summing the atomic masses of all the elements in their molecular formulas. [&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-254905","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/254905","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=254905"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/254905\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=254905"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=254905"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=254905"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}