{"id":192332,"date":"2025-02-17T13:41:19","date_gmt":"2025-02-17T13:41:19","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192332"},"modified":"2025-02-17T13:41:22","modified_gmt":"2025-02-17T13:41:22","slug":"use-the-references-to-access-important-values-if-needed-for-this-question","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/17\/use-the-references-to-access-important-values-if-needed-for-this-question\/","title":{"rendered":"Use the References to access Important values if needed for this question"},"content":{"rendered":"\n

Use the References to access Important values if needed for this question.
Complete the table below for calculating the molar mass of the ionic compound chromium(II) fluoride.
Formula
Molar mass of ion
Number of ions
mol =
mol =
Mass of ion in one mole of chromium(II) fluoride
9
g
Cation
Cr2+
X
g\/mol
Anion
F
X
g\/mol<\/p>\n\n\n\n

Molar mass chromium(II) fluoride = g\/mol<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

To calculate the molar mass of chromium(II) fluoride (CrF\u2082), we need to break it down into its components: the cation (Cr\u00b2\u207a) and the anion (F\u207b). Let\u2019s first identify the molar mass of each ion.<\/p>\n\n\n\n

    \n
  1. Cation: Cr\u00b2\u207a (Chromium ion)<\/strong>
    The atomic mass of chromium (Cr) is approximately 52.00 g\/mol. Since it has a +2 charge (Cr\u00b2\u207a), this value is the same as the atomic mass.<\/li>\n\n\n\n
  2. Anion: F\u207b (Fluoride ion)<\/strong>
    The atomic mass of fluorine (F) is approximately 19.00 g\/mol.<\/li>\n<\/ol>\n\n\n\n

    Now, let\u2019s use this information to fill out the table:<\/p>\n\n\n\n

    Table<\/h3>\n\n\n\n
    Formula<\/strong><\/th>Molar Mass of Ion<\/strong><\/th>Number of Ions<\/strong><\/th>mol =<\/strong><\/th>Mass of Ion in One Mole of Chromium(II) Fluoride<\/strong><\/th><\/tr><\/thead>
    Cation (Cr\u00b2\u207a)<\/strong><\/td>52.00 g\/mol<\/td>2<\/td>mol<\/td>104.00 g\/mol (2 \u00d7 52.00)<\/td><\/tr>
    Anion (F\u207b)<\/strong><\/td>19.00 g\/mol<\/td>2<\/td>mol<\/td>38.00 g\/mol (2 \u00d7 19.00)<\/td><\/tr>
    Molar Mass CrF\u2082<\/strong><\/td><\/td><\/td>mol<\/td>142.00 g\/mol (104.00 + 38.00)<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Explanation:<\/h3>\n\n\n\n

    To determine the molar mass of chromium(II) fluoride (CrF\u2082), we need to consider the mass of each individual ion and how many ions are present in one formula unit of CrF\u2082.<\/p>\n\n\n\n

      \n
    • Chromium(II) fluoride has one Cr\u00b2\u207a ion and two F\u207b ions in each formula unit (CrF\u2082).<\/li>\n\n\n\n
    • The molar mass of Cr\u00b2\u207a is 52.00 g\/mol.<\/li>\n\n\n\n
    • The molar mass of F\u207b is 19.00 g\/mol.<\/li>\n<\/ul>\n\n\n\n

      To find the molar mass of CrF\u2082, we multiply the number of each ion by its molar mass:<\/p>\n\n\n\n

        \n
      • The Cr\u00b2\u207a ion contributes 52.00 g\/mol \u00d7 1 = 52.00 g\/mol.<\/li>\n\n\n\n
      • The F\u207b ions contribute 19.00 g\/mol \u00d7 2 = 38.00 g\/mol.<\/li>\n<\/ul>\n\n\n\n

        Summing these gives us:
        [ 52.00 \\, \\text{g\/mol} + 38.00 \\, \\text{g\/mol} = 90.00 \\, \\text{g\/mol} ]<\/p>\n\n\n\n

        Thus, the molar mass of chromium(II) fluoride is 142.00 g\/mol<\/strong>.<\/p>\n\n\n\n

        Let me generate the table image for you!<\/p>\n\n\n\n

        It looks like you’ve hit your daily maximum number of images. To ensure the best experience for everyone, we have rate limits in place. Please wait for the next day before generating more images. Your daily maximum will reset in 10 hours and 19 minutes.<\/p>\n\n\n\n

        In the meantime, I can still help clarify any questions you have or offer more details if needed!<\/p>\n","protected":false},"excerpt":{"rendered":"

        Use the References to access Important values if needed for this question.Complete the table below for calculating the molar mass of the ionic compound chromium(II) fluoride.FormulaMolar mass of ionNumber of ionsmol =mol =Mass of ion in one mole of chromium(II) fluoride9gCationCr2+Xg\/molAnionFXg\/mol Molar mass chromium(II) fluoride = g\/mol The Correct Answer and Explanation is : To […]<\/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-192332","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192332","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=192332"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192332\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192332"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192332"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192332"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}