{"id":190361,"date":"2025-02-12T07:52:17","date_gmt":"2025-02-12T07:52:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190361"},"modified":"2025-02-12T07:52:20","modified_gmt":"2025-02-12T07:52:20","slug":"if-87-grams-of-k2so4-molar-mass-174-grams-is-dissolved-in-enough-water-to-make-250-ml-of-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/12\/if-87-grams-of-k2so4-molar-mass-174-grams-is-dissolved-in-enough-water-to-make-250-ml-of-solution\/","title":{"rendered":"If 87 grams of K2SO4 (molar mass 174 grams) is dissolved in enough water to make 250 mL of solution"},"content":{"rendered":"\n
    \n
  1. If 87 grams of K2SO4 (molar mass 174 grams) is dissolved in enough water to make 250 mL of solution, what are the concentrations of the potassium and the sulfate ions?
    [K]
    [SO42]<\/li>\n\n\n\n
  2. If 87 grams of K\u00e4, SO4 (molar mass 174 grams) is dissolved in enough water to
    (A)
    0.020 M
    0.020 M
    (B)
    1.0 M
    2.0 M
    (C)
    2.0 M
    1.0 M
    (D)
    2.0 M
    2.0 M
    (E)
    4.0 M
    2.0 M<\/li>\n<\/ol>\n\n\n\n

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

    To calculate the concentrations of potassium ions (K\u207a) and sulfate ions (SO\u2084\u00b2\u207b), we need to follow these steps:<\/p>\n\n\n\n

    Step 1: Calculate the moles of K\u2082SO\u2084<\/h3>\n\n\n\n

    We are given the mass of K\u2082SO\u2084, which is 87 grams, and its molar mass, which is 174 grams per mole. Using this information, we can calculate the number of moles of K\u2082SO\u2084 using the formula:<\/p>\n\n\n\n

    [
    \\text{Moles of K\u2082SO\u2084} = \\frac{\\text{Mass}}{\\text{Molar mass}} = \\frac{87 \\, \\text{g}}{174 \\, \\text{g\/mol}} = 0.5 \\, \\text{mol}
    ]<\/p>\n\n\n\n

    Step 2: Find the molarity of K\u2082SO\u2084<\/h3>\n\n\n\n

    The volume of the solution is 250 mL, which is 0.250 L. The molarity (M) is calculated as:<\/p>\n\n\n\n

    [
    \\text{Molarity of K\u2082SO\u2084} = \\frac{\\text{Moles of K\u2082SO\u2084}}{\\text{Volume of solution in L}} = \\frac{0.5 \\, \\text{mol}}{0.250 \\, \\text{L}} = 2.0 \\, \\text{M}
    ]<\/p>\n\n\n\n

    Step 3: Determine the concentration of potassium ions (K\u207a)<\/h3>\n\n\n\n

    In each formula unit of K\u2082SO\u2084, there are two potassium ions (K\u207a). Therefore, the concentration of K\u207a will be twice the molarity of K\u2082SO\u2084:<\/p>\n\n\n\n

    [
    [\\text{K}^+] = 2 \\times 2.0 \\, \\text{M} = 4.0 \\, \\text{M}
    ]<\/p>\n\n\n\n

    Step 4: Determine the concentration of sulfate ions (SO\u2084\u00b2\u207b)<\/h3>\n\n\n\n

    Each formula unit of K\u2082SO\u2084 also contains one sulfate ion (SO\u2084\u00b2\u207b), so the concentration of sulfate ions will be the same as the molarity of K\u2082SO\u2084:<\/p>\n\n\n\n

    [
    [\\text{SO\u2084}\u00b2\u207b] = 2.0 \\, \\text{M}
    ]<\/p>\n\n\n\n

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

    The concentrations are:<\/p>\n\n\n\n

      \n
    • Potassium ion concentration: 4.0 M<\/strong><\/li>\n\n\n\n
    • Sulfate ion concentration: 2.0 M<\/strong><\/li>\n<\/ul>\n\n\n\n

      Thus, the correct answer is (E)<\/strong>:<\/p>\n\n\n\n

        \n
      • [K\u207a] = 4.0 M<\/li>\n\n\n\n
      • [SO\u2084\u00b2\u207b] = 2.0 M<\/li>\n<\/ul>\n\n\n\n

        I will also generate an image to visualize the steps involved.<\/p>\n\n\n\n

        Here is a visual representation of the calculation for the ion concentrations in the solution. It shows the steps for determining the moles, molarity, and final concentrations of potassium and sulfate ions in K\u2082SO\u2084. Let me know if you need further explanations or details!<\/p>\n\n\n\n

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

        The Correct Answer and Explanation is : To calculate the concentrations of potassium ions (K\u207a) and sulfate ions (SO\u2084\u00b2\u207b), we need to follow these steps: Step 1: Calculate the moles of K\u2082SO\u2084 We are given the mass of K\u2082SO\u2084, which is 87 grams, and its molar mass, which is 174 grams per mole. Using this […]<\/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-190361","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190361","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=190361"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190361\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190361"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190361"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190361"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}