{"id":237406,"date":"2025-06-17T08:08:30","date_gmt":"2025-06-17T08:08:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=237406"},"modified":"2025-06-17T08:08:32","modified_gmt":"2025-06-17T08:08:32","slug":"an-aqueous-solution-of-concentrated-hydrobromic-acid-contains-48-hbr-by-mass","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/17\/an-aqueous-solution-of-concentrated-hydrobromic-acid-contains-48-hbr-by-mass\/","title":{"rendered":"An aqueous solution of concentrated hydrobromic acid contains 48% HBr by mass"},"content":{"rendered":"\n

An aqueous solution of concentrated hydrobromic acid contains 48% HBr by mass. If the density of the solution is 1.5 g\/mL , What is its concentration?<\/p>\n\n\n\n

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

Correct Answer:<\/strong><\/p>\n\n\n\n

To find the concentration in molarity (mol\/L)<\/strong> of the hydrobromic acid (HBr) solution:<\/p>\n\n\n\n

Step 1: Use the definition of percent by mass<\/strong><\/h3>\n\n\n\n

The solution is 48% HBr by mass<\/strong>, which means:<\/p>\n\n\n\n

    \n
  • 48 grams of HBr are present in every 100 grams<\/strong> of solution.<\/li>\n<\/ul>\n\n\n\n

    Step 2: Use the density to find volume<\/strong><\/h3>\n\n\n\n

    The density<\/strong> is given as 1.5 g\/mL<\/strong>, which means:<\/p>\n\n\n\n

      \n
    • 100 grams of solution has a volume of:<\/li>\n<\/ul>\n\n\n\n

      Volume=massdensity=100\u2009g1.5\u2009g\/mL=66.67\u2009mL=0.06667\u2009L\\text{Volume} = \\frac{\\text{mass}}{\\text{density}} = \\frac{100\\, \\text{g}}{1.5\\, \\text{g\/mL}} = 66.67\\, \\text{mL} = 0.06667\\, \\text{L}Volume=densitymass\u200b=1.5g\/mL100g\u200b=66.67mL=0.06667L<\/p>\n\n\n\n

      Step 3: Convert grams of HBr to moles<\/strong><\/h3>\n\n\n\n

      Molar mass of HBr = 1.008 (H) + 79.904 (Br) = 80.912 g\/mol<\/strong>Moles of HBr=48\u2009g80.912\u2009g\/mol\u22480.5935\u2009mol\\text{Moles of HBr} = \\frac{48\\, \\text{g}}{80.912\\, \\text{g\/mol}} \\approx 0.5935\\, \\text{mol}Moles of HBr=80.912g\/mol48g\u200b\u22480.5935mol<\/p>\n\n\n\n

      Step 4: Calculate molarity<\/strong><\/h3>\n\n\n\n

      Molarity=moles of soluteliters of solution=0.5935\u2009mol0.06667\u2009L\u22488.90\u2009mol\/L\\text{Molarity} = \\frac{\\text{moles of solute}}{\\text{liters of solution}} = \\frac{0.5935\\, \\text{mol}}{0.06667\\, \\text{L}} \\approx 8.90\\, \\text{mol\/L}Molarity=liters of solutionmoles of solute\u200b=0.06667L0.5935mol\u200b\u22488.90mol\/L<\/p>\n\n\n\n

      \n\n\n\n

      Final Answer:<\/strong><\/h3>\n\n\n\n

      The concentration of the solution is approximately 8.90 mol\/L<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Explanation (300 words)<\/strong><\/h3>\n\n\n\n

      To determine the molar concentration of a hydrobromic acid solution, we begin with the given data: the solution is 48% HBr by mass and has a density of 1.5 g\/mL. The percentage by mass tells us that in every 100 grams of solution, 48 grams are pure HBr.<\/p>\n\n\n\n

      Next, we convert this mass of solution into volume using the given density. Since density equals mass divided by volume, rearranging the formula gives volume equals mass divided by density. Plugging in the values, 100 grams divided by 1.5 g\/mL gives a volume of 66.67 mL. This is then converted into liters for molarity calculation, giving 0.06667 L.<\/p>\n\n\n\n

      To find how many moles are in the 48 grams of HBr, we use its molar mass. Hydrogen has a mass of approximately 1.008 g\/mol and bromine approximately 79.904 g\/mol, summing to a molar mass of 80.912 g\/mol for HBr. Dividing the mass of HBr (48 g) by its molar mass gives roughly 0.5935 moles.<\/p>\n\n\n\n

      Finally, molarity is defined as moles of solute per liter of solution. Dividing the number of moles (0.5935 mol) by the volume in liters (0.06667 L) yields a concentration of approximately 8.90 mol\/L. This represents the molarity of the hydrobromic acid in the solution.<\/p>\n\n\n\n

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

      An aqueous solution of concentrated hydrobromic acid contains 48% HBr by mass. If the density of the solution is 1.5 g\/mL , What is its concentration? The Correct Answer and Explanation is: Correct Answer: To find the concentration in molarity (mol\/L) of the hydrobromic acid (HBr) solution: Step 1: Use the definition of percent by […]<\/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-237406","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237406","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=237406"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237406\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=237406"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=237406"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=237406"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}