{"id":244919,"date":"2025-07-05T13:41:42","date_gmt":"2025-07-05T13:41:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244919"},"modified":"2025-07-05T13:41:44","modified_gmt":"2025-07-05T13:41:44","slug":"the-molecular-weight-of-ethanol-c2h5oh-is-46-1-and-the-density-of-absolute-100-ethanol-0-79-g-ml","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/the-molecular-weight-of-ethanol-c2h5oh-is-46-1-and-the-density-of-absolute-100-ethanol-0-79-g-ml\/","title":{"rendered":"The molecular weight of ethanol (C2H5OH) is 46.1 and the density of absolute (100%) ethanol = 0.79 g\/mL"},"content":{"rendered":"\n

The molecular weight of ethanol (C2H5OH) is 46.1 and the density of absolute (100%) ethanol = 0.79 g\/mL. a) Calculate the molarity of absolute ethanol (100% ethanol). b) The legal limit for a driver’s blood alcohol in Australia is 50 mg of ethanol per 100 mL of blood (referred to as a blood alcohol content (BAC) of 0.05%). What is the molarity of ethanol (in blood) in a person at this legal limit? c) D’Arenberg The Old Bloke & Three Young Blondes is an expensive (~$200 per bottle) Shiraz dominant wine that has an alcohol content of 14.2% ethanol by volume. Calculate the molarity of ethanol in this wine. d) How much D’Arenberg The Old Bloke & Three Young Blondes (standard serving size = 100 mL) could an 80 kg person drink and remain under the legal limit? The blood alcohol content will be equivalent to the alcohol content in the total ‘water’ component of the human body. Assume that an 80 kg person contains 48 L of ‘water’. Ignore the metabolism, and assume the water content of the person remains constant. e) During a romantic candlelight dinner, an 80 kg person drinks a bottle (750 mL) of D’Arenberg The Old Bloke & Three Young Blondes in the space of 3 hours. Ethanol is metabolized at a constant rate of 120 mg per hour per kg body weight, regardless of its concentration. i. What is their blood alcohol content (mg\/100 mL blood) at the end of the dinner? ii. How long will it take for their blood alcohol level to fall below the legal limit (0.05%)? Postscript: Fortunately, their partner was the designated driver. He\/she enjoyed the meal but only drank water and drove the couple home safely, soundly, and legally.<\/p>\n\n\n\n

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

Let’s break down and solve each part step by step.<\/p>\n\n\n\n

a) Molarity of Absolute Ethanol<\/h3>\n\n\n\n

We know the molecular weight of ethanol is 46.1 g\/mol, and the density is 0.79 g\/mL.<\/p>\n\n\n\n

    \n
  1. Density (\u03c1)<\/strong> = 0.79 g\/mL, meaning 1 mL of ethanol weighs 0.79 grams.<\/li>\n\n\n\n
  2. Molecular weight of ethanol<\/strong> = 46.1 g\/mol, so 46.1 grams of ethanol corresponds to 1 mole.<\/li>\n<\/ol>\n\n\n\n

    To find the molarity (M), we need to convert the mass of ethanol per volume (density) to moles per liter.Molarity=Mass of ethanol in grams per literMolecular weight in grams per mole\\text{Molarity} = \\frac{\\text{Mass of ethanol in grams per liter}}{\\text{Molecular weight in grams per mole}} Molarity=Molecular weight in grams per moleMass of ethanol in grams per liter\u200b<\/p>\n\n\n\n

      \n
    • First, we convert 1 mL of ethanol to grams:
      1\u00a0mL\u00d70.79\u00a0g\/mL=0.79\u00a0g1 \\text{ mL} \\times 0.79 \\text{ g\/mL} = 0.79 \\text{ g}1\u00a0mL\u00d70.79\u00a0g\/mL=0.79\u00a0g.<\/li>\n\n\n\n
    • For 1 L of ethanol (1000 mL):
      1000\u00a0mL\u00d70.79\u00a0g\/mL=790\u00a0g1000 \\text{ mL} \\times 0.79 \\text{ g\/mL} = 790 \\text{ g}1000\u00a0mL\u00d70.79\u00a0g\/mL=790\u00a0g.<\/li>\n\n\n\n
    • Now, calculate moles in 790 g:
      790\u00a0g46.1\u00a0g\/mol\u224817.14\u00a0moles\\frac{790 \\text{ g}}{46.1 \\text{ g\/mol}} \\approx 17.14 \\text{ moles}46.1\u00a0g\/mol790\u00a0g\u200b\u224817.14\u00a0moles.<\/li>\n<\/ul>\n\n\n\n

      Thus, the molarity of absolute ethanol is approximately 17.14 M.<\/p>\n\n\n\n

      \n\n\n\n

      b) Molarity of Ethanol in Blood at Legal Limit<\/h3>\n\n\n\n

      The legal limit for blood alcohol content (BAC) in Australia is 50 mg of ethanol per 100 mL of blood.<\/p>\n\n\n\n

        \n
      1. Amount of ethanol in blood<\/strong> = 50 mg per 100 mL = 0.05 g per 100 mL.<\/li>\n\n\n\n
      2. Convert to grams per liter:
        0.05\u00a0g\/100\u00a0mL=0.5\u00a0g\/L0.05 \\text{ g\/100 mL} = 0.5 \\text{ g\/L}0.05\u00a0g\/100\u00a0mL=0.5\u00a0g\/L.<\/li>\n<\/ol>\n\n\n\n

        Now, we convert this to molarity:Molarity=0.5 g46.1 g\/mol\u22480.0108 M.\\text{Molarity} = \\frac{0.5 \\text{ g}}{46.1 \\text{ g\/mol}} \\approx 0.0108 \\text{ M}.Molarity=46.1 g\/mol0.5 g\u200b\u22480.0108 M.<\/p>\n\n\n\n

        So, the molarity of ethanol in blood at this legal limit is approximately 0.0108 M.<\/p>\n\n\n\n

        \n\n\n\n

        c) Molarity of Ethanol in Wine<\/h3>\n\n\n\n

        The ethanol content of the wine is 14.2% by volume, which means 14.2 mL of ethanol per 100 mL of wine.<\/p>\n\n\n\n

          \n
        1. First, calculate the mass of ethanol in 14.2 mL of wine:\n
            \n
          • The density of ethanol is 0.79 g\/mL, so 14.2\u00a0mL\u00d70.79\u00a0g\/mL=11.21\u00a0g14.2 \\text{ mL} \\times 0.79 \\text{ g\/mL} = 11.21 \\text{ g}14.2\u00a0mL\u00d70.79\u00a0g\/mL=11.21\u00a0g.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Now, calculate the moles of ethanol:<\/li>\n<\/ol>\n\n\n\n

            Moles of ethanol=11.21 g46.1 g\/mol\u22480.243 mol.\\text{Moles of ethanol} = \\frac{11.21 \\text{ g}}{46.1 \\text{ g\/mol}} \\approx 0.243 \\text{ mol}.Moles of ethanol=46.1 g\/mol11.21 g\u200b\u22480.243 mol.<\/p>\n\n\n\n

              \n
            1. Finally, calculate the molarity of ethanol in 100 mL of wine:<\/li>\n<\/ol>\n\n\n\n

              Molarity=0.243 mol0.1 L=2.43 M.\\text{Molarity} = \\frac{0.243 \\text{ mol}}{0.1 \\text{ L}} = 2.43 \\text{ M}.Molarity=0.1 L0.243 mol\u200b=2.43 M.<\/p>\n\n\n\n

              So, the molarity of ethanol in the wine is approximately 2.43 M.<\/p>\n\n\n\n

              \n\n\n\n

              d) Maximum Amount of Wine an 80 kg Person Can Drink<\/h3>\n\n\n\n

              The legal BAC limit is 0.05%. For an 80 kg person, this is equivalent to 50 mg of ethanol per 100 mL of blood.<\/p>\n\n\n\n

              We know:<\/p>\n\n\n\n

                \n
              • The person has 48 L of water in their body.<\/li>\n\n\n\n
              • We need to find how much wine they can drink without exceeding the legal limit.<\/li>\n<\/ul>\n\n\n\n
                  \n
                1. Amount of ethanol the person can have<\/strong>:
                  50 mg\/100 mL is 0.5 g\/L, so for 48 L of blood, the person can have:
                  0.5\u00a0g\/L\u00d748\u00a0L=24\u00a0g\u00a0of\u00a0ethanol0.5 \\text{ g\/L} \\times 48 \\text{ L} = 24 \\text{ g of ethanol}0.5\u00a0g\/L\u00d748\u00a0L=24\u00a0g\u00a0of\u00a0ethanol.<\/li>\n\n\n\n
                2. Ethanol in 100 mL of wine<\/strong>:
                  From part (c), 100 mL of wine contains 11.21 g of ethanol.<\/li>\n<\/ol>\n\n\n\n

                  Now, calculate how much wine they can drink:24 g11.21 g\/100 mL\u22482.14 bottles (of 750 mL).\\frac{24 \\text{ g}}{11.21 \\text{ g\/100 mL}} \\approx 2.14 \\text{ bottles (of 750 mL)}.11.21 g\/100 mL24 g\u200b\u22482.14 bottles (of 750 mL).<\/p>\n\n\n\n

                  Thus, the person can drink approximately 2.14 bottles of 750 mL wine, which is about 1.6 liters of wine, to stay under the legal limit.<\/p>\n\n\n\n

                  \n\n\n\n

                  e) Ethanol Metabolism During Dinner<\/h3>\n\n\n\n

                  Let’s assume the person drinks 750 mL of wine, which contains ethanol at 14.2% by volume.<\/p>\n\n\n\n

                    \n
                  1. Ethanol in 750 mL of wine<\/strong>: 750\u00a0mL\u00d70.142\u00a0(ethanol\u00a0by\u00a0volume)=106.5\u00a0mL\u00a0of\u00a0ethanol.750 \\text{ mL} \\times 0.142 \\text{ (ethanol by volume)} = 106.5 \\text{ mL of ethanol}.750\u00a0mL\u00d70.142\u00a0(ethanol\u00a0by\u00a0volume)=106.5\u00a0mL\u00a0of\u00a0ethanol. Convert this to grams:
                    106.5\u00a0mL\u00d70.79\u00a0g\/mL=84.2\u00a0g106.5 \\text{ mL} \\times 0.79 \\text{ g\/mL} = 84.2 \\text{ g}106.5\u00a0mL\u00d70.79\u00a0g\/mL=84.2\u00a0g.<\/li>\n\n\n\n
                  2. Metabolism Rate<\/strong>:
                    The metabolism rate is 120 mg per hour per kg of body weight. For an 80 kg person:
                    120\u00a0mg\u00d780\u00a0kg=9600\u00a0mg\/hour=9.6\u00a0g\/hour120 \\text{ mg} \\times 80 \\text{ kg} = 9600 \\text{ mg\/hour} = 9.6 \\text{ g\/hour}120\u00a0mg\u00d780\u00a0kg=9600\u00a0mg\/hour=9.6\u00a0g\/hour.<\/li>\n<\/ol>\n\n\n\n

                    Now, let\u2019s answer the two questions:<\/p>\n\n\n\n

                    \n\n\n\n

                    i. Blood Alcohol Content After Dinner<\/h4>\n\n\n\n

                    Time for drinking = 3 hours.<\/p>\n\n\n\n

                    The total ethanol metabolized over 3 hours:9.6 g\/hour\u00d73 hours=28.8 g.9.6 \\text{ g\/hour} \\times 3 \\text{ hours} = 28.8 \\text{ g}.9.6 g\/hour\u00d73 hours=28.8 g.<\/p>\n\n\n\n

                    The person consumes 84.2 g, so the amount left in the body:84.2 g\u221228.8 g=55.4 g.84.2 \\text{ g} – 28.8 \\text{ g} = 55.4 \\text{ g}.84.2 g\u221228.8 g=55.4 g.<\/p>\n\n\n\n

                    Now, calculate the BAC:55.4 g48 L=1.16 g\/L.\\frac{55.4 \\text{ g}}{48 \\text{ L}} = 1.16 \\text{ g\/L}.48 L55.4 g\u200b=1.16 g\/L.<\/p>\n\n\n\n

                    Convert to mg\/100 mL:1.16 g\/L=116 mg\/100 mL.1.16 \\text{ g\/L} = 116 \\text{ mg\/100 mL}.1.16 g\/L=116 mg\/100 mL.<\/p>\n\n\n\n

                    So, the BAC is 116 mg\/100 mL after dinner.<\/p>\n\n\n\n

                    \n\n\n\n

                    ii. Time to Fall Below Legal Limit<\/h4>\n\n\n\n

                    The person\u2019s BAC needs to fall below 50 mg\/100 mL. They have 116 mg\/100 mL at the end of dinner. The metabolism rate is 9600 mg\/hour.<\/p>\n\n\n\n

                    Time required to reduce to the legal limit:116 mg\/100 mL\u221250 mg\/100 mL9600 mg\/hour=669600 hours\u22486.88 hours.\\frac{116 \\text{ mg\/100 mL} – 50 \\text{ mg\/100 mL}}{9600 \\text{ mg\/hour}} = \\frac{66}{9600} \\text{ hours} \\approx 6.88 \\text{ hours}.9600 mg\/hour116 mg\/100 mL\u221250 mg\/100 mL\u200b=960066\u200b hours\u22486.88 hours.<\/p>\n\n\n\n

                    It will take approximately 7 hours<\/strong> for their BAC to fall below the legal limit.<\/p>\n\n\n\n

                    \n\n\n\n

                    This gives us a thorough understanding of the alcohol dynamics in the body in different situations!<\/p>\n\n\n\n

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

                    The molecular weight of ethanol (C2H5OH) is 46.1 and the density of absolute (100%) ethanol = 0.79 g\/mL. a) Calculate the molarity of absolute ethanol (100% ethanol). b) The legal limit for a driver’s blood alcohol in Australia is 50 mg of ethanol per 100 mL of blood (referred to as a blood alcohol content […]<\/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-244919","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244919","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=244919"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244919\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244919"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244919"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244919"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}