{"id":192894,"date":"2025-02-18T10:19:10","date_gmt":"2025-02-18T10:19:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192894"},"modified":"2025-02-18T10:19:13","modified_gmt":"2025-02-18T10:19:13","slug":"calculate-the-mass-of-naoh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/18\/calculate-the-mass-of-naoh\/","title":{"rendered":"Calculate the mass of NaOH"},"content":{"rendered":"\n

1- Calculate the mass of NaOH which is required to prepare 0.1 M in 250 cm3 aqueous solution (molar mass = 40 gm\/mole) 2- Calculate the mass of HCl which is required to prepare 0.1 N in 500 ml aqueous solution. (molar mass = 36.5 gm\/mole) 3- 10 gm KCl dissolved in 400 cm3 aqueous solution. Calculate the molarity and molality of the solution. (molar mass, KCl = 74.5 gm\/mole \u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u0153 density of the solution = 1.1 gm\/ cm3 ). PROBLEMS 4) Calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198 g of water. The molar mass of sulfuric acid is 98.09 g. 5) The density of a 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g\/mL. What is the molality of the solution? The molar mass of methanol is 32.04 g. 6) Calculate the molality of a 35.4 percent (by mass) aqueous solution of phosphoric acid (H3PO4 ). The molar mass of phosphoric acid is 97.99 g 7) A solution containing 53 g of sodium carbonate (mol mass = 106) in 1 Kg water at 20\u00c3\u201a\u00c2\u00b0C. The density of the solution at this temperature is 1.1 g\/ml. Determine for the solution. 1) the mole fraction of each component 1) the mass percent 2) the molarity 3) the molality 4) the normality.<\/p>\n\n\n\n

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

1) Mass of NaOH required to prepare 0.1 M in 250 cm\u00b3 aqueous solution (Molar mass = 40 g\/mol)<\/strong><\/h3>\n\n\n\n

Molarity (M)<\/strong> is defined as moles of solute per liter of solution. To calculate the mass of NaOH, use the formula:<\/p>\n\n\n\n

[
\\text{Moles of NaOH} = M \\times V
]<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • ( M = 0.1 \\, \\text{M} )<\/li>\n\n\n\n
  • ( V = 250 \\, \\text{cm}^3 = 0.25 \\, \\text{L} )<\/li>\n<\/ul>\n\n\n\n

    [
    \\text{Moles of NaOH} = 0.1 \\, \\text{mol\/L} \\times 0.25 \\, \\text{L} = 0.025 \\, \\text{mol}
    ]<\/p>\n\n\n\n

    Now, to find the mass:<\/p>\n\n\n\n

    [
    \\text{Mass of NaOH} = \\text{Moles of NaOH} \\times \\text{Molar mass of NaOH}
    ]<\/p>\n\n\n\n

    [
    \\text{Mass of NaOH} = 0.025 \\, \\text{mol} \\times 40 \\, \\text{g\/mol} = 1 \\, \\text{g}
    ]<\/p>\n\n\n\n

    So, 1 gram of NaOH is required.<\/p>\n\n\n\n

    \n\n\n\n

    2) Mass of HCl required to prepare 0.1 N in 500 mL aqueous solution (Molar mass = 36.5 g\/mol)<\/strong><\/h3>\n\n\n\n

    Normality (N)<\/strong> is defined as the number of equivalents per liter of solution. For HCl, 1 mole of HCl provides 1 equivalent, so normality is the same as molarity.<\/p>\n\n\n\n

    We need the mass of HCl to prepare 0.1 N in 500 mL (0.5 L) solution:<\/p>\n\n\n\n

    [
    \\text{Equivalents of HCl} = N \\times V = 0.1 \\, \\text{eq\/L} \\times 0.5 \\, \\text{L} = 0.05 \\, \\text{eq}
    ]<\/p>\n\n\n\n

    Now, the mass:<\/p>\n\n\n\n

    [
    \\text{Mass of HCl} = \\text{Equivalents of HCl} \\times \\text{Molar mass of HCl}
    ]<\/p>\n\n\n\n

    [
    \\text{Mass of HCl} = 0.05 \\, \\text{eq} \\times 36.5 \\, \\text{g\/mol} = 1.825 \\, \\text{g}
    ]<\/p>\n\n\n\n

    So, 1.825 grams of HCl is required.<\/p>\n\n\n\n

    \n\n\n\n

    3) Molarity and Molality of KCl solution<\/strong><\/h3>\n\n\n\n

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

      \n
    • Mass of KCl = 10 g<\/li>\n\n\n\n
    • Volume of solution = 400 cm\u00b3 = 0.4 L<\/li>\n\n\n\n
    • Molar mass of KCl = 74.5 g\/mol<\/li>\n\n\n\n
    • Density = 1.1 g\/cm\u00b3<\/li>\n<\/ul>\n\n\n\n

      Molarity (M):<\/h4>\n\n\n\n

      [
      \\text{Moles of KCl} = \\frac{\\text{Mass of KCl}}{\\text{Molar mass of KCl}} = \\frac{10 \\, \\text{g}}{74.5 \\, \\text{g\/mol}} = 0.134 \\, \\text{mol}
      ]<\/p>\n\n\n\n

      [
      \\text{Molarity} = \\frac{\\text{Moles of KCl}}{\\text{Volume of solution (in L)}} = \\frac{0.134 \\, \\text{mol}}{0.4 \\, \\text{L}} = 0.335 \\, \\text{M}
      ]<\/p>\n\n\n\n

      Molality (m):<\/h4>\n\n\n\n

      Molality is moles of solute per kilogram of solvent. First, we need to find the mass of the solvent (water). The total mass of the solution is:<\/p>\n\n\n\n

      [
      \\text{Mass of solution} = \\text{Density} \\times \\text{Volume of solution} = 1.1 \\, \\text{g\/cm}^3 \\times 400 \\, \\text{cm}^3 = 440 \\, \\text{g}
      ]<\/p>\n\n\n\n

      Mass of water (solvent):<\/p>\n\n\n\n

      [
      \\text{Mass of water} = 440 \\, \\text{g} – 10 \\, \\text{g} = 430 \\, \\text{g} = 0.430 \\, \\text{kg}
      ]<\/p>\n\n\n\n

      Now, calculate the molality:<\/p>\n\n\n\n

      [
      \\text{Molality} = \\frac{\\text{Moles of KCl}}{\\text{Mass of solvent (in kg)}} = \\frac{0.134 \\, \\text{mol}}{0.430 \\, \\text{kg}} = 0.312 \\, \\text{mol\/kg}
      ]<\/p>\n\n\n\n

      \n\n\n\n

      4) Molality of sulfuric acid solution<\/strong><\/h3>\n\n\n\n

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

        \n
      • Mass of sulfuric acid = 24.4 g<\/li>\n\n\n\n
      • Mass of water = 198 g<\/li>\n\n\n\n
      • Molar mass of sulfuric acid = 98.09 g\/mol<\/li>\n<\/ul>\n\n\n\n

        First, calculate the moles of sulfuric acid:<\/p>\n\n\n\n

        [
        \\text{Moles of H}_2\\text{SO}_4 = \\frac{24.4 \\, \\text{g}}{98.09 \\, \\text{g\/mol}} = 0.249 \\, \\text{mol}
        ]<\/p>\n\n\n\n

        Now, calculate the molality:<\/p>\n\n\n\n

        [
        \\text{Molality} = \\frac{0.249 \\, \\text{mol}}{0.198 \\, \\text{kg}} = 1.26 \\, \\text{mol\/kg}
        ]<\/p>\n\n\n\n

        \n\n\n\n

        5) Molality of a 2.45 M aqueous solution of methanol<\/strong><\/h3>\n\n\n\n

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

          \n
        • Molarity = 2.45 M<\/li>\n\n\n\n
        • Density = 0.976 g\/mL<\/li>\n\n\n\n
        • Molar mass of methanol = 32.04 g\/mol<\/li>\n<\/ul>\n\n\n\n

          First, calculate the mass of 1 L of solution:<\/p>\n\n\n\n

          [
          \\text{Mass of solution} = \\text{Density} \\times \\text{Volume} = 0.976 \\, \\text{g\/mL} \\times 1000 \\, \\text{mL} = 976 \\, \\text{g}
          ]<\/p>\n\n\n\n

          Now, calculate the moles of methanol in 1 L:<\/p>\n\n\n\n

          [
          \\text{Moles of methanol} = 2.45 \\, \\text{mol\/L} \\times 1 \\, \\text{L} = 2.45 \\, \\text{mol}
          ]<\/p>\n\n\n\n

          Mass of methanol:<\/p>\n\n\n\n

          [
          \\text{Mass of methanol} = 2.45 \\, \\text{mol} \\times 32.04 \\, \\text{g\/mol} = 78.5 \\, \\text{g}
          ]<\/p>\n\n\n\n

          Now, mass of water:<\/p>\n\n\n\n

          [
          \\text{Mass of water} = 976 \\, \\text{g} – 78.5 \\, \\text{g} = 897.5 \\, \\text{g} = 0.8975 \\, \\text{kg}
          ]<\/p>\n\n\n\n

          Finally, calculate the molality:<\/p>\n\n\n\n

          [
          \\text{Molality} = \\frac{2.45 \\, \\text{mol}}{0.8975 \\, \\text{kg}} = 2.73 \\, \\text{mol\/kg}
          ]<\/p>\n\n\n\n

          \n\n\n\n

          6) Molality of a 35.4% (by mass) aqueous solution of phosphoric acid<\/strong><\/h3>\n\n\n\n

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

            \n
          • Mass percent of H\u2083PO\u2084 = 35.4%<\/li>\n\n\n\n
          • Mass of solution = 100 g (for simplicity)<\/li>\n\n\n\n
          • Mass of H\u2083PO\u2084 = 35.4 g<\/li>\n\n\n\n
          • Molar mass of H\u2083PO\u2084 = 97.99 g\/mol<\/li>\n<\/ul>\n\n\n\n

            Mass of water:<\/p>\n\n\n\n

            [
            \\text{Mass of water} = 100 \\, \\text{g} – 35.4 \\, \\text{g} = 64.6 \\, \\text{g} = 0.0646 \\, \\text{kg}
            ]<\/p>\n\n\n\n

            Moles of H\u2083PO\u2084:<\/p>\n\n\n\n

            [
            \\text{Moles of H\u2083PO\u2084} = \\frac{35.4 \\, \\text{g}}{97.99 \\, \\text{g\/mol}} = 0.361 \\, \\text{mol}
            ]<\/p>\n\n\n\n

            Molality:<\/p>\n\n\n\n

            [
            \\text{Molality} = \\frac{0.361 \\, \\text{mol}}{0.0646 \\, \\text{kg}} = 5.59 \\, \\text{mol\/kg}
            ]<\/p>\n\n\n\n

            \n\n\n\n

            7) Solution containing 53 g of sodium carbonate (Na\u2082CO\u2083) in 1 kg of water<\/strong><\/h3>\n\n\n\n

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

              \n
            • Mass of Na\u2082CO\u2083 = 53 g<\/li>\n\n\n\n
            • Mass of water = 1 kg<\/li>\n\n\n\n
            • Molar mass of Na\u2082CO\u2083 = 106 g\/mol<\/li>\n\n\n\n
            • Density of solution = 1.1 g\/mL<\/li>\n<\/ul>\n\n\n\n

              Mole fraction:<\/h4>\n\n\n\n

              [
              \\text{Moles of Na\u2082CO\u2083} = \\frac{53 \\, \\text{g}}{106 \\, \\text{g\/mol}} = 0.5 \\, \\text{mol}
              ]<\/p>\n\n\n\n

              Total mass of the solution:<\/p>\n\n\n\n

              [
              \\text{Mass of solution} = 53 \\, \\text{g} + 1000 \\, \\text{g} = 1053 \\, \\text{g}
              ]<\/p>\n\n\n\n

              Moles of water:<\/p>\n\n\n\n

              [
              \\text{Moles of water} = \\frac{1000 \\, \\text{g}}{18 \\, \\text{g\/mol}} = 55.56 \\, \\text{mol}
              ]<\/p>\n\n\n\n

              Mole fraction of Na\u2082CO\u2083:<\/p>\n\n\n\n

              [
              \\chi_{\\text{Na\u2082CO\u2083}} = \\frac{0.5}{0.5 + 55.56} = 0.00896
              ]<\/p>\n\n\n\n

              Mole fraction of water:<\/p>\n\n\n\n

              [
              \\chi_{\\text{H\u2082O}} = \\frac{55.56}{0.5 + 55.56} = 0.991
              ]<\/p>\n\n\n\n

              Mass percent of Na\u2082CO\u2083:<\/h4>\n\n\n\n

              [
              \\text{Mass percent of Na\u2082CO\u2083} = \\frac{53 \\, \\text{g}}{1053 \\, \\text{g}} \\times 100 = 5.03\\%
              ]<\/p>\n\n\n\n

              Molarity:<\/h4>\n\n\n\n

              [
              \\text{Molarity} = \\frac{0.5 \\, \\text{mol}}{1 \\, \\text{L}} = 0.5 \\, \\text{M}
              ]<\/p>\n\n\n\n

              Molality:<\/h4>\n\n\n\n

              [
              \\text{Molality} = \\frac{0.5 \\, \\text{mol}}{1 \\, \\text{kg}} = 0.5 \\, \\text{mol\/kg}
              ]<\/p>\n\n\n\n

              Normality:<\/h4>\n\n\n\n

              [
              \\text{Normality} = 0.5 \\, \\text{N} \\quad (\\text{since Na\u2082CO\u2083 provides 2 equivalents of Na+ ions})
              ]<\/p>\n","protected":false},"excerpt":{"rendered":"

              1- Calculate the mass of NaOH which is required to prepare 0.1 M in 250 cm3 aqueous solution (molar mass = 40 gm\/mole) 2- Calculate the mass of HCl which is required to prepare 0.1 N in 500 ml aqueous solution. (molar mass = 36.5 gm\/mole) 3- 10 gm KCl dissolved in 400 cm3 aqueous […]<\/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-192894","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192894","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=192894"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192894\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192894"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192894"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192894"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}