{"id":192699,"date":"2025-02-18T06:29:13","date_gmt":"2025-02-18T06:29:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192699"},"modified":"2025-02-18T06:29:15","modified_gmt":"2025-02-18T06:29:15","slug":"the-compound-sucrose-is-commonly-known-as-table-sugar","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/18\/the-compound-sucrose-is-commonly-known-as-table-sugar\/","title":{"rendered":"The compound sucrose is commonly known as table sugar"},"content":{"rendered":"\n<p>The compound sucrose is commonly known as table sugar. The chemical formula of sucrose is C12H22011-<br>How many hydrogen atoms are present in 55.7 mg of sucrose? (Avogadro&#8217;s number is N_{0} = 6.022e + 23 )<br>a) There are 4.05e + 21 hydrogen atoms present.<br>b) There are 2.16e + 27 hydrogen atoms present.<br>c) There are 4.45e + 18 hydrogen atoms present.<br>d) There are 2.16e + 21 hydrogen atoms present.<br>e) There are 2.16e + 24 hydrogen atoms present.<\/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<p>To solve the problem, we need to determine how many hydrogen atoms are present in 55.7 mg of sucrose (C\u2081\u2082H\u2082\u2082O\u2081\u2081).<\/p>\n\n\n\n<p><strong>Step 1: Determine the molar mass of sucrose.<\/strong><\/p>\n\n\n\n<p>The molar mass of sucrose is the sum of the atomic masses of all atoms in its formula:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 12.01 g\/mol<\/li>\n\n\n\n<li>Hydrogen (H): 1.008 g\/mol<\/li>\n\n\n\n<li>Oxygen (O): 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>The molar mass of sucrose, C\u2081\u2082H\u2082\u2082O\u2081\u2081, is calculated as:<\/p>\n\n\n\n<p>[<br>(12 \\times 12.01) + (22 \\times 1.008) + (11 \\times 16.00) = 144.12 + 22.176 + 176.00 = 342.30 \\, \\text{g\/mol}<br>]<\/p>\n\n\n\n<p><strong>Step 2: Calculate the moles of sucrose in 55.7 mg.<\/strong><\/p>\n\n\n\n<p>We need to convert the mass of sucrose from milligrams to grams and then calculate the number of moles:<\/p>\n\n\n\n<p>[<br>55.7 \\, \\text{mg} = 0.0557 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>Now, calculate the number of moles:<\/p>\n\n\n\n<p>[<br>\\text{moles of sucrose} = \\frac{\\text{mass of sucrose}}{\\text{molar mass}} = \\frac{0.0557}{342.30} = 1.63 \\times 10^{-4} \\, \\text{mol}<br>]<\/p>\n\n\n\n<p><strong>Step 3: Determine the number of hydrogen atoms in the sucrose sample.<\/strong><\/p>\n\n\n\n<p>From the chemical formula C\u2081\u2082H\u2082\u2082O\u2081\u2081, we see that there are 22 hydrogen atoms in one molecule of sucrose. Using Avogadro\u2019s number (N\u2080 = 6.022 \u00d7 10\u00b2\u00b3 molecules\/mol), we can calculate the number of molecules in 1.63 \u00d7 10\u207b\u2074 moles of sucrose:<\/p>\n\n\n\n<p>[<br>\\text{Number of molecules} = 1.63 \\times 10^{-4} \\, \\text{mol} \\times 6.022 \\times 10^{23} = 9.83 \\times 10^{19} \\, \\text{molecules}<br>]<\/p>\n\n\n\n<p>Since each molecule contains 22 hydrogen atoms, the total number of hydrogen atoms is:<\/p>\n\n\n\n<p>[<br>\\text{Number of hydrogen atoms} = 9.83 \\times 10^{19} \\times 22 = 2.16 \\times 10^{21} \\, \\text{hydrogen atoms}<br>]<\/p>\n\n\n\n<p><strong>Step 4: Conclusion<\/strong><\/p>\n\n\n\n<p>The correct answer is:<\/p>\n\n\n\n<p><strong>d) There are 2.16 \u00d7 10\u00b2\u00b9 hydrogen atoms present.<\/strong><\/p>\n\n\n\n<p>Now, I\u2019ll generate a diagram to illustrate this process.<\/p>\n\n\n\n<p>Here\u2019s a detailed illustration of the process for calculating the number of hydrogen atoms in sucrose, which includes the necessary steps and formulas. I hope it helps visualize the solution!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-895.png\" alt=\"\" class=\"wp-image-192700\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The compound sucrose is commonly known as table sugar. The chemical formula of sucrose is C12H22011-How many hydrogen atoms are present in 55.7 mg of sucrose? (Avogadro&#8217;s number is N_{0} = 6.022e + 23 )a) There are 4.05e + 21 hydrogen atoms present.b) There are 2.16e + 27 hydrogen atoms present.c) There are 4.45e + [&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-192699","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192699","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=192699"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192699\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192699"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192699"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192699"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}