{"id":223778,"date":"2025-06-02T10:40:28","date_gmt":"2025-06-02T10:40:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223778"},"modified":"2025-06-02T10:40:30","modified_gmt":"2025-06-02T10:40:30","slug":"how-to-calculate-normality-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/how-to-calculate-normality-2\/","title":{"rendered":"How to calculate normality"},"content":{"rendered":"\n

How to calculate normality? : 37% HCl solution<\/p>\n\n\n\n

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

How to Calculate Normality of 37% HCl Solution<\/strong><\/h3>\n\n\n\n

To calculate the normality (N)<\/strong> of a 37% HCl solution, follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Step-by-Step Calculation:<\/strong><\/h3>\n\n\n\n
    \n
  1. Identify key information:<\/strong>\n
      \n
    • % by weight (w\/w)<\/strong> = 37% HCl \u2192 37 g HCl in 100 g solution<\/li>\n\n\n\n
    • Density of 37% HCl<\/strong> \u2248 1.19 g\/mL<\/strong><\/li>\n\n\n\n
    • Molar mass of HCl<\/strong> = 36.46 g\/mol<\/li>\n\n\n\n
    • HCl is a monoprotic acid<\/strong>, meaning 1 mole HCl = 1 equivalent<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n
        \n
      1. Calculate volume of 100 g of solution:<\/strong><\/li>\n<\/ol>\n\n\n\n

        Volume=MassDensity=100\u2009g1.19\u2009g\/mL\u224884.03\u2009mL=0.08403\u2009L\\text{Volume} = \\frac{\\text{Mass}}{\\text{Density}} = \\frac{100\\, \\text{g}}{1.19\\, \\text{g\/mL}} \\approx 84.03\\, \\text{mL} = 0.08403\\, \\text{L}<\/p>\n\n\n\n

        \n\n\n\n
          \n
        1. Calculate moles of HCl:<\/strong><\/li>\n<\/ol>\n\n\n\n

          Moles of HCl=37\u2009g36.46\u2009g\/mol\u22481.0148\u2009mol\\text{Moles of HCl} = \\frac{37\\, \\text{g}}{36.46\\, \\text{g\/mol}} \\approx 1.0148\\, \\text{mol}<\/p>\n\n\n\n

          \n\n\n\n
            \n
          1. Calculate normality (N):<\/strong><\/li>\n<\/ol>\n\n\n\n

            Since HCl provides 1 equivalent per mole: Normality (N)=Equivalents of HClVolume in L=1.01480.08403\u224812.07\u2009N\\text{Normality (N)} = \\frac{\\text{Equivalents of HCl}}{\\text{Volume in L}} = \\frac{1.0148}{0.08403} \\approx \\boxed{12.07\\, \\text{N}}<\/p>\n\n\n\n

            \n\n\n\n

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

            Normality (N)<\/strong> is a concentration unit that reflects the number of equivalents of a solute per liter of solution. It’s particularly useful for acids and bases where reactivity depends on the number of ionizable hydrogen or hydroxide ions. For hydrochloric acid (HCl), each molecule donates one proton (H\u207a), making it a monoprotic acid<\/strong>. Therefore, 1 mole = 1 equivalent<\/strong>, and the normality is numerically equal to molarity for HCl.<\/p>\n\n\n\n

            In a 37% (w\/w) solution, 37 g of pure HCl is present in 100 g of total solution. To determine how concentrated this solution is in terms of normality, we need to express this in terms of equivalents per liter. The first step is converting 100 g of solution to volume using the density<\/strong> of 37% HCl, which is approximately 1.19 g\/mL<\/strong>. This gives a volume of about 84.03 mL<\/strong>, or 0.08403 L<\/strong>.<\/p>\n\n\n\n

            Then, calculate how many moles of HCl are in 37 g using its molar mass (36.46 g\/mol)<\/strong>, resulting in approximately 1.0148 moles<\/strong>, and because HCl is monoprotic, this also equals 1.0148 equivalents<\/strong>.<\/p>\n\n\n\n

            Finally, dividing equivalents by volume in liters gives a normality of ~12.07 N<\/strong>. This tells us that each liter of this concentrated HCl solution can provide 12.07 equivalents of hydrogen ions, making it very strong and corrosive.<\/p>\n\n\n\n

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

            How to calculate normality? : 37% HCl solution The Correct Answer and Explanation is: How to Calculate Normality of 37% HCl Solution To calculate the normality (N) of a 37% HCl solution, follow these steps: Step-by-Step Calculation: Volume=MassDensity=100\u2009g1.19\u2009g\/mL\u224884.03\u2009mL=0.08403\u2009L\\text{Volume} = \\frac{\\text{Mass}}{\\text{Density}} = \\frac{100\\, \\text{g}}{1.19\\, \\text{g\/mL}} \\approx 84.03\\, \\text{mL} = 0.08403\\, \\text{L} Moles of HCl=37\u2009g36.46\u2009g\/mol\u22481.0148\u2009mol\\text{Moles of HCl} = \\frac{37\\, […]<\/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-223778","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223778","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=223778"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223778\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223778"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223778"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223778"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}