{"id":266784,"date":"2025-07-23T07:12:11","date_gmt":"2025-07-23T07:12:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=266784"},"modified":"2025-07-23T07:12:15","modified_gmt":"2025-07-23T07:12:15","slug":"concentrated-hcl-has-a-specific-gravity-of-1-18-and-contains-37-by-weight-hcl","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/23\/concentrated-hcl-has-a-specific-gravity-of-1-18-and-contains-37-by-weight-hcl\/","title":{"rendered":"Concentrated HCl has a specific gravity of 1.18 and contains 37% by weight HCl."},"content":{"rendered":"\n

Concentrated HCl has a specific gravity of 1.18 and contains 37% by weight HCl. Specific gravity is interpreted as density in grams per milliliter. Calculate the molarity of concentrated HCl, and determine how much concentrated acid should be diluted to 1 L to make 0.1 M HCl.<\/p>\n\n\n\n

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

To calculate the molarity of concentrated HCl, we need to break down the information and apply a few key steps:<\/p>\n\n\n\n

Step 1: Determine the density of concentrated HCl.<\/h3>\n\n\n\n

The specific gravity of concentrated HCl is given as 1.18, which means the density is 1.18 g\/mL.<\/p>\n\n\n\n

Step 2: Find the mass of HCl in 1 liter of concentrated acid.<\/h3>\n\n\n\n

We know the density is 1.18 g\/mL, so the mass of 1 liter (1000 mL) of concentrated HCl is:Mass of solution=Density\u00d7Volume\\text{Mass of solution} = \\text{Density} \\times \\text{Volume}Mass of solution=Density\u00d7VolumeMass of solution=1.18\u2009g\/mL\u00d71000\u2009mL=1180\u2009g\\text{Mass of solution} = 1.18 \\, \\text{g\/mL} \\times 1000 \\, \\text{mL} = 1180 \\, \\text{g}Mass of solution=1.18g\/mL\u00d71000mL=1180g<\/p>\n\n\n\n

Next, since the solution is 37% by weight HCl, the mass of HCl in 1180 grams of solution is:Mass of HCl=37%\u00d71180\u2009g=0.37\u00d71180\u2009g=436.6\u2009g\\text{Mass of HCl} = 37\\% \\times 1180 \\, \\text{g} = 0.37 \\times 1180 \\, \\text{g} = 436.6 \\, \\text{g}Mass of HCl=37%\u00d71180g=0.37\u00d71180g=436.6g<\/p>\n\n\n\n

Step 3: Convert the mass of HCl to moles.<\/h3>\n\n\n\n

The molar mass of HCl is approximately 36.46 g\/mol. To find the moles of HCl in the solution:Moles of HCl=Mass of HClMolar mass of HCl\\text{Moles of HCl} = \\frac{\\text{Mass of HCl}}{\\text{Molar mass of HCl}}Moles of HCl=Molar mass of HClMass of HCl\u200bMoles of HCl=436.6\u2009g36.46\u2009g\/mol\u224811.97\u2009mol\\text{Moles of HCl} = \\frac{436.6 \\, \\text{g}}{36.46 \\, \\text{g\/mol}} \\approx 11.97 \\, \\text{mol}Moles of HCl=36.46g\/mol436.6g\u200b\u224811.97mol<\/p>\n\n\n\n

Step 4: Calculate the molarity of the concentrated HCl.<\/h3>\n\n\n\n

Molarity (M) is defined as the number of moles of solute per liter of solution. Since we have 1 liter of solution:Molarity of concentrated HCl=11.97\u2009mol1\u2009L\u224812.0\u2009M\\text{Molarity of concentrated HCl} = \\frac{11.97 \\, \\text{mol}}{1 \\, \\text{L}} \\approx 12.0 \\, \\text{M}Molarity of concentrated HCl=1L11.97mol\u200b\u224812.0M<\/p>\n\n\n\n

Thus, the molarity of the concentrated HCl is approximately 12.0 M.<\/p>\n\n\n\n

Step 5: Determine the volume of concentrated acid needed to make 0.1 M HCl.<\/h3>\n\n\n\n

To dilute the concentrated HCl to a 0.1 M solution, we use the dilution equation:M1V1=M2V2M_1V_1 = M_2V_2M1\u200bV1\u200b=M2\u200bV2\u200b<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • M1=12.0\u2009MM_1 = 12.0 \\, \\text{M}M1\u200b=12.0M (initial molarity),<\/li>\n\n\n\n
  • V1V_1V1\u200b is the volume of concentrated acid required,<\/li>\n\n\n\n
  • M2=0.1\u2009MM_2 = 0.1 \\, \\text{M}M2\u200b=0.1M (final molarity),<\/li>\n\n\n\n
  • V2=1\u2009LV_2 = 1 \\, \\text{L}V2\u200b=1L (final volume).<\/li>\n<\/ul>\n\n\n\n

    Now, solve for V1V_1V1\u200b:12.0\u2009M\u00d7V1=0.1\u2009M\u00d71\u2009L12.0 \\, \\text{M} \\times V_1 = 0.1 \\, \\text{M} \\times 1 \\, \\text{L}12.0M\u00d7V1\u200b=0.1M\u00d71LV1=0.1\u2009M\u00d71\u2009L12.0\u2009M=0.112\u2009L=0.00833\u2009LV_1 = \\frac{0.1 \\, \\text{M} \\times 1 \\, \\text{L}}{12.0 \\, \\text{M}} = \\frac{0.1}{12} \\, \\text{L} = 0.00833 \\, \\text{L}V1\u200b=12.0M0.1M\u00d71L\u200b=120.1\u200bL=0.00833LV1\u22488.33\u2009mLV_1 \\approx 8.33 \\, \\text{mL}V1\u200b\u22488.33mL<\/p>\n\n\n\n

    Final Answer:<\/h3>\n\n\n\n
      \n
    • The molarity of concentrated HCl is 12.0 M<\/strong>.<\/li>\n\n\n\n
    • To make 1 liter of 0.1 M HCl, you will need to dilute 8.33 mL<\/strong> of the concentrated HCl solution.<\/li>\n<\/ul>\n\n\n\n
      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Concentrated HCl has a specific gravity of 1.18 and contains 37% by weight HCl. Specific gravity is interpreted as density in grams per milliliter. Calculate the molarity of concentrated HCl, and determine how much concentrated acid should be diluted to 1 L to make 0.1 M HCl. The Correct Answer and Explanation is: To calculate […]<\/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-266784","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/266784","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=266784"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/266784\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=266784"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=266784"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=266784"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}