{"id":221903,"date":"2025-05-30T17:23:07","date_gmt":"2025-05-30T17:23:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221903"},"modified":"2025-05-30T17:23:10","modified_gmt":"2025-05-30T17:23:10","slug":"calculate-the-mass-of-vanadiumv-oxide-v2o5-that-contains-a-million-1-000-x-106-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-million-1-000-x-106-vanadium-atoms\/","title":{"rendered":"Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.000 \u00d7 106) vanadium atoms."},"content":{"rendered":"\n

Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.000 \u00d7 106) vanadium atoms. Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct Answer:<\/strong><\/h3>\n\n\n\n

The mass of vanadium(V) oxide (V\u2082O\u2085<\/strong>) that contains 1.000 \u00d7 10\u2076 vanadium atoms<\/strong> is:<\/p>\n\n\n\n

2.517 \u00d7 10\u207b\u00b9\u2077 g<\/strong><\/p>\n\n\n\n

\n\n\n\n

Detailed Explanation <\/strong><\/h3>\n\n\n\n

To calculate the mass of vanadium(V) oxide (V\u2082O\u2085)<\/strong> containing 1.000 \u00d7 10\u2076 vanadium atoms<\/strong>, we can follow a step-by-step method using stoichiometry, atomic masses, and Avogadro\u2019s number.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Understand the Composition of V\u2082O\u2085<\/strong><\/h4>\n\n\n\n
    \n
  • The chemical formula V\u2082O\u2085<\/strong> tells us that:\n
      \n
    • 1 mole of V\u2082O\u2085<\/strong> contains 2 moles of vanadium (V)<\/strong> atoms.<\/li>\n\n\n\n
    • 1 mole of V\u2082O\u2085<\/strong> contains 5 moles of oxygen (O)<\/strong> atoms.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

      So, if we know the number of vanadium atoms, we can determine how many moles of V\u2082O\u2085<\/strong> they belong to.<\/p>\n\n\n\n

      \n\n\n\n

      Step 2: Use Avogadro\u2019s Number<\/strong><\/h4>\n\n\n\n

      Avogadro\u2019s number = 6.022 \u00d7 10\u00b2\u00b3 atoms\/mol<\/strong><\/p>\n\n\n\n

      So, the number of moles of vanadium atoms is: mol V=1.000\u00d7106 atoms6.022\u00d71023 atoms\/mol=1.661\u00d710\u221218 mol\\text{mol V} = \\frac{1.000 \\times 10^6 \\text{ atoms}}{6.022 \\times 10^{23} \\text{ atoms\/mol}} = 1.661 \\times 10^{-18} \\text{ mol}<\/p>\n\n\n\n

      Since each mole of V\u2082O\u2085<\/strong> contains 2 mol V<\/strong>, the number of moles of V\u2082O\u2085<\/strong> is: mol V\u2082O\u2085=1.661\u00d710\u2212182=8.306\u00d710\u221219 mol\\text{mol V\u2082O\u2085} = \\frac{1.661 \\times 10^{-18}}{2} = 8.306 \\times 10^{-19} \\text{ mol}<\/p>\n\n\n\n

      \n\n\n\n

      Step 3: Molar Mass of V\u2082O\u2085<\/strong><\/h4>\n\n\n\n
        \n
      • Atomic mass of V<\/strong> = 50.94 g\/mol<\/li>\n\n\n\n
      • Atomic mass of O<\/strong> = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n

        So, molar mass of V\u2082O\u2085: (2\u00d750.94)+(5\u00d716.00)=101.88+80.00=181.88 g\/mol(2 \\times 50.94) + (5 \\times 16.00) = 101.88 + 80.00 = 181.88 \\text{ g\/mol}<\/p>\n\n\n\n

        \n\n\n\n

        Step 4: Calculate Mass<\/strong><\/h4>\n\n\n\n

        mass=mol V\u2082O\u2085\u00d7molar mass=(8.306\u00d710\u221219)\u00d7181.88=2.517\u00d710\u221217 g\\text{mass} = \\text{mol V\u2082O\u2085} \\times \\text{molar mass} = (8.306 \\times 10^{-19}) \\times 181.88 = 2.517 \\times 10^{-17} \\text{ g}<\/p>\n\n\n\n

        \n\n\n\n

        Final Answer:<\/strong><\/h3>\n\n\n\n

        2.517 \u00d7 10\u207b\u00b9\u2077 g of V\u2082O\u2085<\/strong> contains 1.000 \u00d7 10\u2076 vanadium atoms.<\/strong><\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.000 \u00d7 106) vanadium atoms. Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits. The Correct Answer and Explanation is: Correct Answer: The mass of vanadium(V) oxide (V\u2082O\u2085) that contains 1.000 \u00d7 10\u2076 vanadium atoms is: […]<\/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-221903","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221903","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=221903"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221903\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221903"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221903"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221903"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}