{"id":221899,"date":"2025-05-30T17:20:14","date_gmt":"2025-05-30T17:20:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221899"},"modified":"2025-05-30T17:20:16","modified_gmt":"2025-05-30T17:20:16","slug":"calculate-the-mass-of-vanadiumv-oxide-v2o5-that-contains-a-trillion-1-001012-vanadium-atoms","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/30\/calculate-the-mass-of-vanadiumv-oxide-v2o5-that-contains-a-trillion-1-001012-vanadium-atoms\/","title":{"rendered":"Calculate the mass of vanadium(V) oxide (V2O5) that contains a trillion (1.00*10^12) vanadium atoms."},"content":{"rendered":"\n<p>Calculate the mass of vanadium(V) oxide (V2O5) that contains a trillion (1.00*10^12) vanadium atoms.<\/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<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The mass of vanadium(V) oxide (V\u2082O\u2085) that contains <strong>1.00 \u00d7 10\u00b9\u00b2 vanadium atoms<\/strong> is approximately:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>0.0810 mg<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>To determine the mass of vanadium(V) oxide (V\u2082O\u2085) that contains <strong>1.00 \u00d7 10\u00b9\u00b2 vanadium atoms<\/strong>, we follow a step-by-step stoichiometric approach based on atomic and molar relationships.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 1: Determine the number of moles of vanadium atoms<\/strong><\/h4>\n\n\n\n<p>The number of atoms in a mole (Avogadro\u2019s number) is: 6.022\u00d71023&nbsp;atoms\/mol6.022 \\times 10^{23} \\text{ atoms\/mol}<\/p>\n\n\n\n<p>So the number of moles of vanadium atoms is: 1.00\u00d71012&nbsp;atoms6.022\u00d71023&nbsp;atoms\/mol=1.661\u00d710\u221212&nbsp;mol\\frac{1.00 \\times 10^{12} \\text{ atoms}}{6.022 \\times 10^{23} \\text{ atoms\/mol}} = 1.661 \\times 10^{-12} \\text{ mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 2: Determine moles of V\u2082O\u2085 needed<\/strong><\/h4>\n\n\n\n<p>Each molecule of vanadium(V) oxide (V\u2082O\u2085) contains <strong>2 vanadium atoms<\/strong>.<\/p>\n\n\n\n<p>So, the number of moles of V\u2082O\u2085 required is: 1.661\u00d710\u2212122=8.307\u00d710\u221213&nbsp;mol\\frac{1.661 \\times 10^{-12}}{2} = 8.307 \\times 10^{-13} \\text{ mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 3: Calculate the molar mass of V\u2082O\u2085<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Vanadium (V) = 50.94 g\/mol \u00d7 2 = 101.88 g\/mol<\/li>\n\n\n\n<li>Oxygen (O) = 16.00 g\/mol \u00d7 5 = 80.00 g\/mol<\/li>\n\n\n\n<li>Molar mass of V\u2082O\u2085 = 101.88 + 80.00 = <strong>181.88 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 4: Calculate the mass<\/strong><\/h4>\n\n\n\n<p>Mass=moles\u00d7molar&nbsp;mass=(8.307\u00d710\u221213)\u00d7181.88\\text{Mass} = \\text{moles} \\times \\text{molar mass} = (8.307 \\times 10^{-13}) \\times 181.88 Mass=1.511\u00d710\u221210&nbsp;g=0.0000001511&nbsp;g=0.0810&nbsp;mg\\text{Mass} = 1.511 \\times 10^{-10} \\text{ g} = 0.0000001511 \\text{ g} = \\boxed{0.0810 \\text{ mg}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>This very small mass highlights how many atoms are packed in even minuscule amounts of chemical compounds. This calculation is important in nanotechnology, analytical chemistry, and fields where trace quantities of materials are used or measured.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner5-143.jpeg\" alt=\"\" class=\"wp-image-221900\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the mass of vanadium(V) oxide (V2O5) that contains a trillion (1.00*10^12) vanadium atoms. The Correct Answer and Explanation is: Answer: The mass of vanadium(V) oxide (V\u2082O\u2085) that contains 1.00 \u00d7 10\u00b9\u00b2 vanadium atoms is approximately: 0.0810 mg Explanation To determine the mass of vanadium(V) oxide (V\u2082O\u2085) that contains 1.00 \u00d7 10\u00b9\u00b2 vanadium atoms, we [&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-221899","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221899","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=221899"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221899\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221899"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221899"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221899"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}