{"id":228185,"date":"2025-06-07T06:26:18","date_gmt":"2025-06-07T06:26:18","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=228185"},"modified":"2025-06-07T06:26:21","modified_gmt":"2025-06-07T06:26:21","slug":"a-solution-contains-448-mg-of-kci-mw74-5-and-468-mg-of-nacl-61-mw58-5-in-500-ml","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/07\/a-solution-contains-448-mg-of-kci-mw74-5-and-468-mg-of-nacl-61-mw58-5-in-500-ml\/","title":{"rendered":"A solution contains 448 mg of KCI (MW=74.5) and 468 mg of NaCl -61 (MW=58.5) in 500 ml"},"content":{"rendered":"\n

A solution contains 448 mg of KCI (MW=74.5) and 468 mg of NaCl -61 (MW=58.5) in 500 ml. Which of the following is the osmolar concentration of ?this solution A. 0.056 B. 0.065 C. 1.556 D. 1.256<\/p>\n\n\n\n

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

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • KCl:<\/strong> 448 mg<\/li>\n\n\n\n
  • NaCl:<\/strong> 468 mg<\/li>\n\n\n\n
  • Molecular weight (MW)<\/strong> of KCl = 74.5 g\/mol<\/li>\n\n\n\n
  • MW of NaCl<\/strong> = 58.5 g\/mol<\/li>\n\n\n\n
  • Volume of solution<\/strong> = 500 mL = 0.5 L<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step 1: Convert mg to grams<\/strong><\/h3>\n\n\n\n
      \n
    • KCl: 448 mg = 0.448 g<\/li>\n\n\n\n
    • NaCl: 468 mg = 0.468 g<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 2: Calculate moles of each solute<\/strong><\/h3>\n\n\n\n

      Moles of KCl=0.44874.5\u22480.00601 mol\\text{Moles of KCl} = \\frac{0.448}{74.5} \\approx 0.00601 \\text{ mol}Moles of KCl=74.50.448\u200b\u22480.00601 molMoles of NaCl=0.46858.5\u22480.008 mol\\text{Moles of NaCl} = \\frac{0.468}{58.5} \\approx 0.008 \\text{ mol}Moles of NaCl=58.50.468\u200b\u22480.008 mol<\/p>\n\n\n\n

      \n\n\n\n

      Step 3: Determine osmoles<\/strong><\/h3>\n\n\n\n

      Both KCl and NaCl are strong electrolytes that dissociate completely in water:<\/p>\n\n\n\n

        \n
      • KCl \u2192 K\u207a + Cl\u207b \u2192 2 particles<\/li>\n\n\n\n
      • NaCl \u2192 Na\u207a + Cl\u207b \u2192 2 particles<\/li>\n<\/ul>\n\n\n\n

        So, osmoles = moles \u00d7 number of particles<\/strong>:<\/p>\n\n\n\n

          \n
        • Osmoles of KCl = 0.00601 \u00d7 2 = 0.01202<\/li>\n\n\n\n
        • Osmoles of NaCl = 0.008 \u00d7 2 = 0.016<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Step 4: Total osmoles in 0.5 L<\/strong><\/h3>\n\n\n\n

          Total osmoles=0.01202+0.016=0.02802 osmoles\\text{Total osmoles} = 0.01202 + 0.016 = 0.02802 \\text{ osmoles}Total osmoles=0.01202+0.016=0.02802 osmoles<\/p>\n\n\n\n

          \n\n\n\n

          Step 5: Osmolarity (Osmol\/L)<\/strong><\/h3>\n\n\n\n

          Osmolarity=0.028020.5=0.05604 Osmol\/L\\text{Osmolarity} = \\frac{0.02802}{0.5} = 0.05604 \\text{ Osmol\/L}Osmolarity=0.50.02802\u200b=0.05604 Osmol\/L<\/p>\n\n\n\n

          \n\n\n\n

          \u2705 Final Answer: A. 0.056 Osmol\/L<\/strong><\/h3>\n\n\n\n
          \n\n\n\n

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

          Osmolarity is a measure of the total concentration of dissolved particles (osmoles)<\/strong> in a solution. In biological and chemical contexts, it reflects how many solute particles are present per liter of solvent. This is important in understanding osmotic pressure and how substances move across membranes.<\/p>\n\n\n\n

          In this problem, we are given the masses of two ionic compounds: potassium chloride (KCl) and sodium chloride (NaCl). These are typical salts that fully dissociate into two ions each in aqueous solution\u2014K\u207a and Cl\u207b for KCl, and Na\u207a and Cl\u207b for NaCl. Each mole of either salt produces 2 moles of particles, which doubles their contribution to osmolarity.<\/p>\n\n\n\n

          First, we convert the mass of each salt to grams, then divide by their respective molecular weights to find the number of moles. After that, since each dissociates into two particles, we multiply the moles by 2 to obtain osmoles.<\/p>\n\n\n\n

          The solution volume is 500 mL, which is 0.5 liters. Osmolarity is defined as osmoles per liter, so we divide the total osmoles by 0.5 L to get the final osmolar concentration.<\/p>\n\n\n\n

          The final calculated value is 0.056 Osmol\/L, or 56 mOsmol\/L. This is a relatively dilute solution and would be considered hypotonic compared to normal blood plasma (~285\u2013295 mOsmol\/L). This kind of calculation is crucial in medicine, particularly in intravenous fluid therapy, where precise osmolarity matters for cell hydration and function.<\/p>\n\n\n\n

          Answer: A. 0.056 Osmol\/L<\/strong>.<\/p>\n\n\n\n

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

          A solution contains 448 mg of KCI (MW=74.5) and 468 mg of NaCl -61 (MW=58.5) in 500 ml. Which of the following is the osmolar concentration of ?this solution A. 0.056 B. 0.065 C. 1.556 D. 1.256 The Correct Answer and Explanation is: Given: Step 1: Convert mg to grams Step 2: Calculate moles of […]<\/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-228185","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228185","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=228185"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228185\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=228185"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=228185"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=228185"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}