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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Depression of freezing point of the solvent

  • question_answer
    Calculate the molal depression constant of a solvent which has freezing point \[{{16.6}^{o}}C\] and latent heat of fusion \[180.75\,J{{g}^{-1}}\].      [Orissa JEE  2005]

    A)                 2.68       

    B)                 3.86

    C)                 4.68       

    D)                 2.86t6

    Correct Answer: B

    Solution :

           \[{{K}_{f}}=\frac{RT_{f}^{2}}{1000\times {{L}_{f}}}\], \[R=8.314J{{K}^{-1}}mo{{l}^{-1}}\]               \[\,{{T}_{f}}=273+16.6=289.6K\] ; \[{{L}_{f}}=180.75\,J{{g}^{-1}}\]                         \[{{K}_{f}}=\frac{8.314\times 289.6\times 289.6}{1000\times 180.75}\]


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