{"id":188677,"date":"2025-02-07T08:03:52","date_gmt":"2025-02-07T08:03:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188677"},"modified":"2025-02-07T08:04:09","modified_gmt":"2025-02-07T08:04:09","slug":"show-all-your-work-and-explain-each-procedur","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/07\/show-all-your-work-and-explain-each-procedur\/","title":{"rendered":"Show all your work and explain each procedure"},"content":{"rendered":"\n

Show all your work and explain each procedure.<\/p>\n\n\n\n

A. 1.00 L of 2.00 M aqueous solution of Sodium Hydroxide (NaOH).<\/p>\n\n\n\n

B. 90.0 mL of a 1.20 M aqueous solution of sodium oxalate (Na2 C2 O4) from a 2.00 M solution of Na2 C2 O4<\/p>\n\n\n\n

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

Let’s break down the problem and explain each procedure.<\/p>\n\n\n\n

Part A: Calculating the Amount of Sodium Hydroxide (NaOH)<\/strong><\/h3>\n\n\n\n

We are given a 2.00 M (molar) solution of sodium hydroxide (NaOH), and we need to find out how many moles of NaOH are in a 1.00 L (liter) solution.<\/p>\n\n\n\n

Step 1: Use the molarity formula<\/strong>
The molarity formula is:
[
M = \\frac{\\text{moles of solute}}{\\text{volume of solution (in L)}}
]
Where:<\/p>\n\n\n\n

    \n
  • ( M ) is the molarity of the solution<\/li>\n\n\n\n
  • Moles of solute is the number of moles of NaOH<\/li>\n\n\n\n
  • Volume of solution is the amount of solution in liters<\/li>\n<\/ul>\n\n\n\n

    We know:<\/p>\n\n\n\n

      \n
    • ( M = 2.00 \\, \\text{M} )<\/li>\n\n\n\n
    • Volume of solution = 1.00 L<\/li>\n<\/ul>\n\n\n\n

      To find the moles of NaOH:
      [
      \\text{Moles of NaOH} = M \\times \\text{Volume of solution (in L)}
      ]
      [
      \\text{Moles of NaOH} = 2.00 \\, \\text{mol\/L} \\times 1.00 \\, \\text{L} = 2.00 \\, \\text{mol}
      ]<\/p>\n\n\n\n

      Thus, there are 2.00 moles of NaOH<\/strong> in 1.00 L of the 2.00 M solution.<\/p>\n\n\n\n

      \n\n\n\n

      Part B: Dilution of Sodium Oxalate (Na\u2082C\u2082O\u2084)<\/strong><\/h3>\n\n\n\n

      We are given a 2.00 M solution of sodium oxalate (Na\u2082C\u2082O\u2084), and we need to determine how to prepare 90.0 mL of a 1.20 M solution of sodium oxalate from this stock solution.<\/p>\n\n\n\n

      Step 1: Use the dilution equation<\/strong>
      The dilution equation is:
      [
      M_1 V_1 = M_2 V_2
      ]
      Where:<\/p>\n\n\n\n

        \n
      • ( M_1 ) is the molarity of the concentrated solution (2.00 M)<\/li>\n\n\n\n
      • ( V_1 ) is the volume of the concentrated solution that we need to take<\/li>\n\n\n\n
      • ( M_2 ) is the molarity of the diluted solution (1.20 M)<\/li>\n\n\n\n
      • ( V_2 ) is the volume of the diluted solution (90.0 mL or 0.090 L)<\/li>\n<\/ul>\n\n\n\n

        We need to solve for ( V_1 ) (the volume of the concentrated solution we will use):
        [
        V_1 = \\frac{M_2 \\times V_2}{M_1}
        ]
        Substitute the known values:
        [
        V_1 = \\frac{1.20 \\, \\text{M} \\times 0.0900 \\, \\text{L}}{2.00 \\, \\text{M}} = \\frac{0.108}{2.00} = 0.0540 \\, \\text{L}
        ]
        So, ( V_1 = 0.0540 \\, \\text{L} ), or 54.0 mL.<\/p>\n\n\n\n

        Step 2: Prepare the diluted solution<\/strong>
        To prepare 90.0 mL of 1.20 M sodium oxalate solution:<\/p>\n\n\n\n

          \n
        1. Take 54.0 mL of the 2.00 M sodium oxalate solution.<\/li>\n\n\n\n
        2. Add distilled water to make the total volume 90.0 mL.<\/li>\n<\/ol>\n\n\n\n

          This gives the required concentration of 1.20 M.<\/p>\n\n\n\n

          \n\n\n\n

          Summary of Results:<\/strong><\/h3>\n\n\n\n
            \n
          • In part A, there are 2.00 moles of NaOH<\/strong> in the 1.00 L solution of 2.00 M NaOH.<\/li>\n\n\n\n
          • In part B, to prepare 90.0 mL of a 1.20 M Na\u2082C\u2082O\u2084 solution from a 2.00 M stock solution, you need to use 54.0 mL of the stock solution<\/strong> and dilute it with water to a total volume of 90.0 mL.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

            Show all your work and explain each procedure. A. 1.00 L of 2.00 M aqueous solution of Sodium Hydroxide (NaOH). B. 90.0 mL of a 1.20 M aqueous solution of sodium oxalate (Na2 C2 O4) from a 2.00 M solution of Na2 C2 O4 The Correct Answer and Explanation is : Let’s break down the […]<\/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-188677","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188677","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=188677"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188677\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188677"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188677"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188677"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}