{"id":229750,"date":"2025-06-08T11:59:10","date_gmt":"2025-06-08T11:59:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229750"},"modified":"2025-06-08T11:59:12","modified_gmt":"2025-06-08T11:59:12","slug":"how-many-molecules-are-present-in-4-61-a-10-2-mol-of-o2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/how-many-molecules-are-present-in-4-61-a-10-2-mol-of-o2\/","title":{"rendered":"How many molecules are present in 4.61 \u00c3\u2014 10^(-2) mol of O2"},"content":{"rendered":"\n<p>How many molecules are present in 4.61 \u00c3\u2014 10^(-2) mol of O2<\/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 find the number of molecules in 4.61 \u00d7 10\u207b\u00b2 mol of O\u2082, we use <strong>Avogadro\u2019s number<\/strong>, which is:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p><strong>6.022 \u00d7 10\u00b2\u00b3 molecules\/mol<\/strong><\/p>\n<\/blockquote>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-Step Solution:<\/strong><\/h3>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Amount of oxygen gas (O\u2082) = <strong>4.61 \u00d7 10\u207b\u00b2 mol<\/strong><\/li>\n\n\n\n<li>Avogadro\u2019s number = <strong>6.022 \u00d7 10\u00b2\u00b3 molecules\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>To find the number of <strong>O\u2082 molecules<\/strong>, multiply the number of moles by Avogadro\u2019s number:Number&nbsp;of&nbsp;molecules=(4.61\u00d710\u22122&nbsp;mol)\u00d7(6.022\u00d71023&nbsp;molecules\/mol)\\text{Number of molecules} = (4.61 \\times 10^{-2} \\text{ mol}) \\times (6.022 \\times 10^{23} \\text{ molecules\/mol})Number&nbsp;of&nbsp;molecules=(4.61\u00d710\u22122&nbsp;mol)\u00d7(6.022\u00d71023&nbsp;molecules\/mol)=2.777\u00d71022&nbsp;molecules= 2.777 \\times 10^{22} \\text{ molecules}=2.777\u00d71022&nbsp;molecules<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p><strong>2.78 \u00d7 10\u00b2\u00b2 molecules of O\u2082<\/strong> (rounded to three significant figures)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udcd8 <strong>Explanation<\/strong><\/h3>\n\n\n\n<p>In chemistry, the concept of the &#8220;mole&#8221; is essential for relating microscopic particles like atoms and molecules to measurable quantities in the laboratory. One mole of any substance contains <strong>6.022 \u00d7 10\u00b2\u00b3 representative particles<\/strong>, whether those are atoms, molecules, ions, or electrons. This large number is known as <strong>Avogadro\u2019s number<\/strong>.<\/p>\n\n\n\n<p>In this case, we are dealing with oxygen gas (O\u2082), which is made up of <strong>diatomic molecules<\/strong>\u2014each molecule contains two oxygen atoms. The question asks how many <strong>molecules<\/strong> are present in <strong>4.61 \u00d7 10\u207b\u00b2 moles<\/strong> of oxygen gas.<\/p>\n\n\n\n<p>To determine this, we use a direct multiplication:<br>We take the number of moles and multiply by Avogadro\u2019s number. This gives us the total number of <strong>molecules<\/strong>, not atoms. Since each O\u2082 molecule consists of two oxygen atoms, if the question had asked for <strong>atoms<\/strong>, we would have multiplied the result by 2. But since we are only asked for <strong>molecules<\/strong>, the answer remains at <strong>2.78 \u00d7 10\u00b2\u00b2 molecules<\/strong>.<\/p>\n\n\n\n<p>This type of calculation is foundational in stoichiometry, where chemists use balanced chemical equations to calculate the amounts of reactants and products. Understanding how to convert between moles and number of particles is essential for mastering concepts like gas laws, molar masses, and chemical reactions.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner6-154.jpeg\" alt=\"\" class=\"wp-image-229751\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>How many molecules are present in 4.61 \u00c3\u2014 10^(-2) mol of O2 The Correct Answer and Explanation is: To find the number of molecules in 4.61 \u00d7 10\u207b\u00b2 mol of O\u2082, we use Avogadro\u2019s number, which is: 6.022 \u00d7 10\u00b2\u00b3 molecules\/mol Step-by-Step Solution: We are given: To find the number of O\u2082 molecules, multiply the [&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-229750","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229750","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=229750"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229750\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229750"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229750"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229750"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}