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

  • question_answer
    0.440 g of a substance dissolved in 22.2 g of benzene lowered the freezing point of benzene by \[{{0.567}^{o}}C\]. The molecular mass of the substance \[({{K}_{f}}={{5.12}^{o}}C\,mo{{l}^{-1}})\] [BHU 2001; CPMT 2001]

    A)                 178.9     

    B)                 177.8

    C)                 176.7     

    D)                 175.6

    Correct Answer: A

    Solution :

                \[m=\frac{{{K}_{f}}\times w\times 1000}{\Delta {{T}_{f}}\times W}\]\[=\frac{5.12\times 0.440\times 1000}{0.567\times 22.2}=178.9\]


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