{"id":244961,"date":"2025-07-05T14:36:25","date_gmt":"2025-07-05T14:36:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244961"},"modified":"2025-07-05T14:36:27","modified_gmt":"2025-07-05T14:36:27","slug":"calculate-the-molar-mass-for-crno32","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/calculate-the-molar-mass-for-crno32\/","title":{"rendered":"Calculate the molar mass for Cr(NO3)2"},"content":{"rendered":"\n

Calculate the molar mass for Cr(NO3)2 (show calculation): b. Calculate the number of moles of Cr(NO3)3 contained in 25.0 grams of Cr(NO3)3:<\/p>\n\n\n\n

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

a. Molar mass of Cr(NO3)2:<\/strong><\/p>\n\n\n\n

To calculate the molar mass of Cr(NO3)2, we need to add up the atomic masses of each element in the compound.<\/p>\n\n\n\n

    \n
  1. Chromium (Cr):<\/strong> The atomic mass of chromium is 52.00 g\/mol<\/strong>.<\/li>\n\n\n\n
  2. Nitrogen (N):<\/strong> The atomic mass of nitrogen is 14.01 g\/mol<\/strong>. There are 2 nitrogen atoms<\/strong> in the compound, so the total mass contributed by nitrogen is 2\u00d714.01=28.02\u2009g\/mol2 \\times 14.01 = 28.02 \\, \\text{g\/mol}2\u00d714.01=28.02g\/mol.<\/li>\n\n\n\n
  3. Oxygen (O):<\/strong> The atomic mass of oxygen is 16.00 g\/mol<\/strong>. There are 6 oxygen atoms<\/strong> (since each nitrate group has 3 oxygen atoms, and there are 2 nitrate groups), so the total mass contributed by oxygen is 6\u00d716.00=96.00\u2009g\/mol6 \\times 16.00 = 96.00 \\, \\text{g\/mol}6\u00d716.00=96.00g\/mol.<\/li>\n<\/ol>\n\n\n\n

    Now, we can calculate the molar mass of Cr(NO3)2:Molar mass=52.00\u2009g\/mol (Cr)+28.02\u2009g\/mol (N)+96.00\u2009g\/mol (O)=176.02\u2009g\/mol.\\text{Molar mass} = 52.00 \\, \\text{g\/mol (Cr)} + 28.02 \\, \\text{g\/mol (N)} + 96.00 \\, \\text{g\/mol (O)} = 176.02 \\, \\text{g\/mol}.Molar mass=52.00g\/mol (Cr)+28.02g\/mol (N)+96.00g\/mol (O)=176.02g\/mol.<\/p>\n\n\n\n

    b. Calculate the number of moles of Cr(NO3)3 in 25.0 grams:<\/strong><\/p>\n\n\n\n

    To calculate the number of moles, we use the formula:Moles=Mass (g)Molar mass (g\/mol).\\text{Moles} = \\frac{\\text{Mass (g)}}{\\text{Molar mass (g\/mol)}}.Moles=Molar mass (g\/mol)Mass (g)\u200b.<\/p>\n\n\n\n

    Given:<\/p>\n\n\n\n

      \n
    • Mass = 25.0 grams<\/strong>.<\/li>\n\n\n\n
    • Molar mass of Cr(NO3)3 = (52.00 g\/mol for Cr) + (3 \u00d7 14.01 g\/mol for N) + (9 \u00d7 16.00 g\/mol for O)<\/strong>.<\/li>\n<\/ul>\n\n\n\n

      Molar mass of Cr(NO3)3=52.00+42.03+144.00=238.03\u2009g\/mol.\\text{Molar mass of Cr(NO3)3} = 52.00 + 42.03 + 144.00 = 238.03 \\, \\text{g\/mol}.Molar mass of Cr(NO3)3=52.00+42.03+144.00=238.03g\/mol.<\/p>\n\n\n\n

      Now, calculate the moles:Moles=25.0\u2009g238.03\u2009g\/mol\u22480.105\u2009moles.\\text{Moles} = \\frac{25.0 \\, \\text{g}}{238.03 \\, \\text{g\/mol}} \\approx 0.105 \\, \\text{moles}.Moles=238.03g\/mol25.0g\u200b\u22480.105moles.<\/p>\n\n\n\n

      Final Answers:<\/h3>\n\n\n\n
        \n
      • a. The molar mass of Cr(NO3)2 is 176.02 g\/mol<\/strong>.<\/li>\n\n\n\n
      • b. The number of moles of Cr(NO3)3 in 25.0 grams is approximately 0.105 moles<\/strong>.<\/li>\n<\/ul>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Calculate the molar mass for Cr(NO3)2 (show calculation): b. Calculate the number of moles of Cr(NO3)3 contained in 25.0 grams of Cr(NO3)3: The Correct Answer and Explanation is: a. Molar mass of Cr(NO3)2: To calculate the molar mass of Cr(NO3)2, we need to add up the atomic masses of each element in the compound. Now, […]<\/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-244961","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244961","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=244961"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244961\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244961"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244961"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244961"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}