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AMU Medical AMU Solved Paper-2015

  • question_answer
    How many grams of \[KCl\] should be added to kg of water to lower its freezing point to \[{{8.0}^{o}}C\]?  [\[{{K}_{f}}\] for water \[={{1.86}^{o}}C\,k\,g\,mo{{l}^{-1}}\])

    A)  160.2

    B)                  150.2

    C)  140.2                   

    D)  130.2

    Correct Answer: A

    Solution :

                    \[\Delta {{T}_{f}}=\frac{{{K}_{f}}\times w\times 1000}{m\times W}\] Given,   \[\Delta {{T}_{f}}={{8}^{o}}C\]                 \[{{K}_{f}}={{1.86}^{o}}C\,kg\,mo{{l}^{-1}}\]                 m = mass of solute = 35.5 + 19 = 54.5                 w = 1kg \[\therefore \]  \[8=\frac{1.86\times w\times 100}{54.5\times 1}\]                 \[W=320.5\,g\]                 \[KCl{{K}^{+}}+C{{l}^{-}}\,(i=2)\] Hence,  \[w=\frac{w}{2}=\frac{320.5}{2}=160.2\,g\]


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