{"id":231912,"date":"2025-06-11T14:42:31","date_gmt":"2025-06-11T14:42:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231912"},"modified":"2025-06-11T14:42:33","modified_gmt":"2025-06-11T14:42:33","slug":"sodium-carbonate-has-the-formula-na2co3-10h2o","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/sodium-carbonate-has-the-formula-na2co3-10h2o\/","title":{"rendered":"Sodium Carbonate has the formula Na2CO3.10H2O"},"content":{"rendered":"\n

Sodium Carbonate has the formula Na2CO3.10H2O. Calculate the relative molecular mass of sodium carbonate. Calculate the amount of sodium carbonate required to make 100 cm3 of 0.25 M solution. Accurately weigh the appropriate amount on an electronic balance in a weighing boat. Transfer into a conical flask and add 100 cm3 of deionized water using a 100 cm3 measuring cylinder. Gently swirl the mixture until the sodium carbonate dissolves. Calculate the moles of sodium carbonate you would have in 10 cm3 of a 0.25 M solution.<\/p>\n\n\n\n

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

1. Calculating the Relative Molecular Mass (Mr) of Sodium Carbonate Decahydrate (Na\u2082CO\u2083\u00b710H\u2082O):<\/strong><\/h3>\n\n\n\n

We calculate the relative molecular mass by adding the atomic masses of all atoms in the formula:<\/p>\n\n\n\n

    \n
  • Na (Sodium) = 22.99 \u00d7 2 = 45.98<\/li>\n\n\n\n
  • C (Carbon) = 12.01 \u00d7 1 = 12.01<\/li>\n\n\n\n
  • O (Oxygen in carbonate) = 16.00 \u00d7 3 = 48.00<\/li>\n\n\n\n
  • H\u2082O (Water of crystallization) = (2\u00d71.008 + 16.00) = 18.02 \u00d7 10 = 180.20<\/li>\n<\/ul>\n\n\n\n

    Total Mr = 45.98 + 12.01 + 48.00 + 180.20 = 286.19 g\/mol<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    2. Calculating the Mass Required to Prepare 100 cm\u00b3 of 0.25 M Sodium Carbonate Solution:<\/strong><\/h3>\n\n\n\n

    To calculate the mass, first calculate the number of moles required:<\/p>\n\n\n\n

    Moles = concentration \u00d7 volume (in L)<\/strong>
    = 0.25 mol\/L \u00d7 0.100 L = 0.025 mol<\/strong><\/p>\n\n\n\n

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

    Mass = moles \u00d7 Mr<\/strong>
    = 0.025 mol \u00d7 286.19 g\/mol = 7.15 g<\/strong><\/p>\n\n\n\n

    So, you need 7.15 grams<\/strong> of sodium carbonate decahydrate.<\/p>\n\n\n\n

    \n\n\n\n

    3. Practical Preparation Steps:<\/strong><\/h3>\n\n\n\n
      \n
    1. Weigh 7.15 g<\/strong> of Na\u2082CO\u2083\u00b710H\u2082O accurately on an electronic balance using a weighing boat.<\/li>\n\n\n\n
    2. Transfer the solid into a clean 250 cm\u00b3 conical flask.<\/li>\n\n\n\n
    3. Use a 100 cm\u00b3 measuring cylinder<\/strong> to measure and add exactly 100 cm\u00b3 of deionized water<\/strong>.<\/li>\n\n\n\n
    4. Swirl the flask gently until the sodium carbonate is completely dissolved.<\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      4. Moles of Sodium Carbonate in 10 cm\u00b3 of 0.25 M Solution:<\/strong><\/h3>\n\n\n\n

      Moles = concentration \u00d7 volume (in L)<\/strong>
      = 0.25 mol\/L \u00d7 0.010 L = 0.0025 mol<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Explanation<\/strong><\/h3>\n\n\n\n

      Sodium carbonate decahydrate (Na\u2082CO\u2083\u00b710H\u2082O) is a commonly used salt in titrations and solution preparation. To accurately prepare a standard solution, it is essential to know the relative molecular mass, which accounts for both the anhydrous part and the water of crystallization. For sodium carbonate decahydrate, the total molecular mass is the sum of the atomic masses of all constituent elements, including ten water molecules, which significantly increase the molar mass.<\/p>\n\n\n\n

      To prepare a 0.25 M solution in 100 cm\u00b3 (or 0.1 L) of water, we use the molarity formula:
      Moles = Molarity \u00d7 Volume (L).
      This gives us the number of moles needed, which is then multiplied by the relative molecular mass to calculate the required mass of solute. In this case, 0.025 mol \u00d7 286.19 g\/mol = 7.15 g.<\/p>\n\n\n\n

      This precise mass is then measured using an electronic balance and transferred to a conical flask. Adding 100 cm\u00b3 of deionized water using a measuring cylinder ensures accurate dilution. Gentle swirling helps dissolve the solid fully, producing a clear, homogeneous solution.<\/p>\n\n\n\n

      Understanding the molarity also allows us to determine the amount of substance in any given volume. For instance, a 10 cm\u00b3 sample from this solution contains 0.0025 mol of sodium carbonate, calculated again using the molarity formula. These calculations are fundamental in quantitative chemistry, particularly in titrations where exact concentrations are necessary to determine the composition of unknown solutions.<\/p>\n\n\n\n

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

      Sodium Carbonate has the formula Na2CO3.10H2O. Calculate the relative molecular mass of sodium carbonate. Calculate the amount of sodium carbonate required to make 100 cm3 of 0.25 M solution. Accurately weigh the appropriate amount on an electronic balance in a weighing boat. Transfer into a conical flask and add 100 cm3 of deionized water using […]<\/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-231912","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231912","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=231912"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231912\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231912"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231912"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231912"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}