{"id":226412,"date":"2025-06-05T08:46:53","date_gmt":"2025-06-05T08:46:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226412"},"modified":"2025-06-05T08:46:55","modified_gmt":"2025-06-05T08:46:55","slug":"what-is-the-molarity-of-a-solution-if-44-0-grams-of-solid-sodium-carbonate-is-added-to-enough-water-to-make-250-0-ml-of-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/what-is-the-molarity-of-a-solution-if-44-0-grams-of-solid-sodium-carbonate-is-added-to-enough-water-to-make-250-0-ml-of-solution\/","title":{"rendered":"What is the molarity of a solution if 44.0 grams of solid sodium carbonate is added to enough water to make 250.0 ml of solution"},"content":{"rendered":"\n

What is the molarity of a solution if 44.0 grams of solid sodium carbonate is added to enough water to make 250.0 ml of solution? Molecular mass of Na2CO3=106 gm]<\/p>\n\n\n\n

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

To calculate the molarity<\/strong> of a solution, we use the formula: Molarity (M)=moles of soluteliters of solution\\text{Molarity (M)} = \\frac{\\text{moles of solute}}{\\text{liters of solution}}<\/p>\n\n\n\n

\n\n\n\n

Step 1: Calculate the moles of sodium carbonate (Na\u2082CO\u2083)<\/strong><\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • Mass of Na\u2082CO\u2083 = 44.0 grams<\/strong><\/li>\n\n\n\n
  • Molar mass of Na\u2082CO\u2083 = 106 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n

    Now calculate moles: Moles of Na\u2082CO\u2083=44.0 g106 g\/mol\u22480.4151 mol\\text{Moles of Na\u2082CO\u2083} = \\frac{44.0\\ \\text{g}}{106\\ \\text{g\/mol}} \\approx 0.4151\\ \\text{mol}<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Convert the volume of solution to liters<\/strong><\/h3>\n\n\n\n

    We are given:<\/p>\n\n\n\n

      \n
    • Volume = 250.0 mL<\/strong><\/li>\n<\/ul>\n\n\n\n

      250.0 mL=0.2500 L250.0\\ \\text{mL} = 0.2500\\ \\text{L}<\/p>\n\n\n\n

      \n\n\n\n

      Step 3: Calculate the molarity<\/strong><\/h3>\n\n\n\n

      Molarity=0.4151 mol0.2500 L=1.6604 mol\/L\\text{Molarity} = \\frac{0.4151\\ \\text{mol}}{0.2500\\ \\text{L}} = 1.6604\\ \\text{mol\/L}<\/p>\n\n\n\n

      \n\n\n\n

      \u2705 Final Answer: Molarity = 1.66 M<\/strong> (rounded to 3 significant figures)<\/h3>\n\n\n\n
      \n\n\n\n

      \ud83d\udd0d Explanation )<\/strong><\/h3>\n\n\n\n

      Molarity is a concentration term used in chemistry to express the amount of solute (in moles) present in one liter of solution. It is commonly denoted by M<\/strong> and has the units mol\/L<\/strong>. To find molarity, we need two key pieces of information: the number of moles of solute<\/strong> and the volume of the solution in liters<\/strong>.<\/p>\n\n\n\n

      In this case, the solute is sodium carbonate (Na\u2082CO\u2083)<\/strong>, a white solid salt that dissolves in water. The problem gives us the mass of sodium carbonate<\/strong> (44.0 grams) and the final volume of the solution<\/strong> (250.0 mL).<\/p>\n\n\n\n

      First, we use the molecular mass<\/strong> of sodium carbonate (106 g\/mol) to convert grams to moles. This conversion is critical because molarity is based on moles, not grams. 44.0 g106 g\/mol\u22480.4151 mol\\frac{44.0\\ \\text{g}}{106\\ \\text{g\/mol}} \\approx 0.4151\\ \\text{mol}<\/p>\n\n\n\n

      Next, we convert the volume of the solution from milliliters to liters by dividing by 1000. This ensures consistency with the units required for molarity. 250.0 mL=0.2500 L250.0\\ \\text{mL} = 0.2500\\ \\text{L}<\/p>\n\n\n\n

      Finally, we divide the moles by the volume in liters: 0.41510.2500=1.66 M\\frac{0.4151}{0.2500} = 1.66\\ \\text{M}<\/p>\n\n\n\n

      This means the solution has a concentration of 1.66 moles per liter<\/strong> of sodium carbonate. Molarity is essential in stoichiometry, titration, and reaction calculations, as it directly relates the amount of a substance to the volume it occupies in a solution.<\/p>\n\n\n\n

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

      What is the molarity of a solution if 44.0 grams of solid sodium carbonate is added to enough water to make 250.0 ml of solution? Molecular mass of Na2CO3=106 gm] The Correct Answer and Explanation is: To calculate the molarity of a solution, we use the formula: Molarity (M)=moles of soluteliters of solution\\text{Molarity (M)} = \\frac{\\text{moles of solute}}{\\text{liters of solution}} […]<\/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-226412","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226412","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=226412"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226412\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226412"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226412"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226412"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}