{"id":29489,"date":"2025-06-21T05:07:39","date_gmt":"2025-06-21T05:07:39","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=29489"},"modified":"2025-06-21T05:07:41","modified_gmt":"2025-06-21T05:07:41","slug":"assuming-equal-concentrations-and-complete-dissociation-rank-these-aqueous-solutions-by-their-freezing-points","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/assuming-equal-concentrations-and-complete-dissociation-rank-these-aqueous-solutions-by-their-freezing-points\/","title":{"rendered":"Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points"},"content":{"rendered":"\n

‘Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points_ Highest freezing point Lowes freezing point NHACI Li;PO4 NazSOa’<\/p>\n\n\n\n

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

To determine the freezing points of aqueous solutions of NH\u2084Cl<\/strong>, Li\u2083PO\u2084<\/strong>, and Na\u2082SO\u2084<\/strong>, we need to consider freezing point depression<\/strong>, which depends on the van\u2019t Hoff factor (i)<\/strong>\u2014the number of particles into which a solute dissociates in solution. The more particles produced, the greater the freezing point depression, and thus the lower<\/strong> the freezing point.<\/p>\n\n\n\n

\n\n\n\n

Dissociation of the Compounds:<\/strong><\/h3>\n\n\n\n
    \n
  1. NH\u2084Cl<\/strong> (Ammonium chloride)
    Dissociates into: NH\u2084\u207a + Cl\u207b
    i = 2<\/strong><\/li>\n\n\n\n
  2. Na\u2082SO\u2084<\/strong> (Sodium sulfate)
    Dissociates into: 2Na\u207a + SO\u2084\u00b2\u207b
    i = 3<\/strong><\/li>\n\n\n\n
  3. Li\u2083PO\u2084<\/strong> (Lithium phosphate)
    Dissociates into: 3Li\u207a + PO\u2084\u00b3\u207b
    i = 4<\/strong><\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Freezing Point Depression (\u0394Tf):<\/strong><\/h3>\n\n\n\n

    The freezing point depression is calculated using the formula:<\/p>\n\n\n\n

    \u0394Tf = i \u00d7 Kf \u00d7 m<\/p>\n\n\n\n

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

      \n
    • i<\/strong> is the van\u2019t Hoff factor<\/li>\n\n\n\n
    • Kf<\/strong> is the cryoscopic constant (same for water in all cases)<\/li>\n\n\n\n
    • m<\/strong> is the molality (assumed equal for all solutions)<\/li>\n<\/ul>\n\n\n\n

      Since all solutions have the same concentration and solvent, \u0394Tf is directly proportional to i<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      Ranking (Highest to Lowest Freezing Point):<\/strong><\/h3>\n\n\n\n
        \n
      • NH\u2084Cl<\/strong> (i = 2) \u2192 Highest freezing point<\/strong><\/li>\n\n\n\n
      • Na\u2082SO\u2084<\/strong> (i = 3) \u2192 Intermediate freezing point<\/strong><\/li>\n\n\n\n
      • Li\u2083PO\u2084<\/strong> (i = 4) \u2192 Lowest freezing point<\/strong><\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

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

        Freezing point depression occurs because the presence of solute particles interferes with the formation of a solid structure, requiring a lower temperature to freeze. More dissociated ions lead to a greater effect. Although all solutes are ionic and fully dissociate, they produce different numbers of ions per formula unit. NH\u2084Cl produces 2 ions, Na\u2082SO\u2084 yields 3, and Li\u2083PO\u2084 gives 4. Therefore, Li\u2083PO\u2084 causes the largest depression and has the lowest freezing point. Conversely, NH\u2084Cl causes the least depression, so it has the highest freezing point.<\/p>\n\n\n\n

        \n\n\n\n

        Final order: NH\u2084Cl > Na\u2082SO\u2084 > Li\u2083PO\u2084<\/strong><\/p>\n\n\n\n

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

        ‘Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points_ Highest freezing point Lowes freezing point NHACI Li;PO4 NazSOa’ The Correct Answer and Explanation is: To determine the freezing points of aqueous solutions of NH\u2084Cl, Li\u2083PO\u2084, and Na\u2082SO\u2084, we need to consider freezing point depression, which depends on the van\u2019t Hoff […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-29489","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29489","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=29489"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29489\/revisions"}],"predecessor-version":[{"id":29491,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29489\/revisions\/29491"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=29489"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=29489"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=29489"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}