{"id":148881,"date":"2024-02-27T16:50:33","date_gmt":"2024-02-27T11:20:33","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=148881"},"modified":"2024-02-27T16:59:08","modified_gmt":"2024-02-27T11:29:08","slug":"18-g-of-a-non-volatile-solute-is-dissolved-in-200-g-of-h-freezes-at-272-07","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/quiz\/18-g-of-a-non-volatile-solute-is-dissolved-in-200-g-of-h-freezes-at-272-07","title":{"rendered":"18 g of a non-volatile solute is dissolved in 200 g of H, freezes at 272.07…"},"content":{"rendered":"

18 g of a non-volatile solute is dissolved in 200 g of H, freezes at 272.07 K. Calculate the molecular mass of solute (K, for water = 1.86 K kg mol-1)<\/h2>\n

Ans.<\/h3>\n

To solve this problem, we can use the concept of freezing point depression.<\/h3>\n

The freezing point depression (\u0394<\/span>T<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>) is given by the formula:<\/h3>\n

\u0394<\/span>T<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>K<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u22c5<\/span><\/span>m<\/span><\/span><\/h3>\n

where:<\/h3>\n
    \n
  • \n

    \u0394<\/span>T<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is the freezing point depression,<\/h3>\n<\/li>\n
  • \n

    K<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is the cryoscopic constant (which is 1.86\u2009K\u00a0kg\u00a0mol\u22121<\/span>1.86<\/span>K\u00a0kg\u00a0mol<\/span>\u22121<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> for water),<\/h3>\n<\/li>\n
  • \n

    m is the molality of the solution.<\/h3>\n<\/li>\n<\/ul>\n

    We first need to find the molality (m) of the solution, which is the number of moles of solute per kilogram of solvent.<\/h3>\n

    Given:<\/h3>\n
      \n
    • \n

      Mass of solvent(water), m<\/span>H<\/span>2<\/span>\u200b<\/span>O<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>200<\/span>g<\/span><\/span>=<\/span><\/span>0.2<\/span>kg<\/span><\/span><\/span><\/h3>\n<\/li>\n
    • \n

      Mass of solute, m<\/span>solute<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>18<\/span>g<\/span><\/span><\/span><\/h3>\n<\/li>\n<\/ul>\n

      Molar mass of solvent (water), M<\/span>H<\/span>2<\/span>\u200b<\/span>O<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>18<\/span>g\/mol<\/span><\/span><\/span><\/h3>\n

      Moles of solute, n<\/span>solute<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>M<\/span>solute<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>m<\/span>solute<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

      Molality, m<\/span>=<\/span><\/span>m<\/span>solvent<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>n<\/span>solute<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

      Now, we can calculate the freezing point depression (\u0394<\/span>T<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>) using the given cryoscopic constant and the molality of the solution.<\/h3>\n

      Finally, the freezing point of the solvent is decreased by \u0394<\/span>T<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> from its normal freezing point of 273.15<\/span>K<\/span><\/span><\/span><\/span><\/span><\/span> to 273.15\u2212\u0394\ufffd\ufffd<\/span>273.15<\/span>\u2212<\/span><\/span>\u0394<\/span>T<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> K.<\/h3>\n

      Let’s calculate:<\/h3>\n

      First, find the molality: \ufffdsolute=18\u2009g\ufffdsolute<\/span>n<\/span>solute<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>M<\/span>solute<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>18g<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> \ufffdsolute=18\u2009g18\u2009g\/mol=1\u2009mol<\/span>n<\/span>solute<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>18g\/mol<\/span><\/span><\/span>18g<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>1<\/span>mol<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

      \ufffd=\ufffdsolute\ufffdsolvent=1\u2009mol0.2\u2009kg=5\u2009mol\/kg<\/span>m<\/span>=<\/span><\/span>m<\/span>solvent<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>n<\/span>solute<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>0.2kg<\/span><\/span><\/span>1mol<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>5<\/span>mol\/kg<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

      Now, calculate the freezing point depression: \u0394\ufffd\ufffd=\ufffd\ufffd\u22c5\ufffd=1.86\u2009K\u00a0kg\u00a0mol\u22121\u00d75\u2009mol\/kg=9.3\u2009K<\/span>\u0394<\/span>T<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>K<\/span>f<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u22c5<\/span><\/span>m<\/span>=<\/span><\/span>1.86<\/span>K\u00a0kg\u00a0mol<\/span>\u22121<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>5<\/span>mol\/kg<\/span><\/span>=<\/span><\/span>9.3<\/span>K<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

      The freezing point of the solvent decreases by 9.3\u2009K<\/span>9.3<\/span>K<\/span><\/span><\/span><\/span><\/span><\/span> from its normal freezing point of 273.15\u2009K<\/span>273.15<\/span>K<\/span><\/span><\/span><\/span><\/span><\/span> to 273.15\u22129.3=263.85\u2009K<\/span>273.15<\/span>\u2212<\/span><\/span>9.3<\/span>=<\/span><\/span>263.85<\/span>K<\/span><\/span><\/span><\/span><\/span><\/span>.<\/h3>\n

      Therefore, the molecular mass of the solute is 18\u2009g\/mol<\/span>18<\/span>g\/mol<\/span><\/span><\/span><\/span><\/span><\/span>.<\/h3>\n","protected":false},"excerpt":{"rendered":"

      18 g of a non-volatile solute is dissolved in 200 g of H, freezes at 272.07 K. Calculate the molecular mass of solute (K, for water = 1.86 K kg mol-1) Ans. To solve this problem, we can use the concept of freezing point depression. The freezing point depression (\u0394Tf\u200b) is given by the formula: […]<\/p>\n","protected":false},"author":21830,"featured_media":114041,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12119],"tags":[],"class_list":{"0":"post-148881","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-quiz"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148881","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=148881"}],"version-history":[{"count":3,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148881\/revisions"}],"predecessor-version":[{"id":148888,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148881\/revisions\/148888"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/114041"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=148881"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=148881"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=148881"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}