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BHU PMT BHU PMT Solved Paper-2001

  • question_answer
    If 0.44 g of substance dissolved in 22.2 g of benzene lowers the freezing point of benzene by\[0.567{}^\circ C,\]then the molecular mass of substance is : (the molal depression constant\[={{5.12}^{o}}C\,mo{{l}^{-1}}\])

    A)  128.4                                   

    B)  156.6

    C)  178.9                                   

    D)  232.4

    Correct Answer: C

    Solution :

                     Key Idea: The molecular mass of a substance is calculated by following relationship \[M=\frac{{{K}_{f}}\times w\times 1000}{\Delta {{T}_{f}}\times W}\] Where, M = molecular weight of solute = ? \[{{K}_{f}}=\]molal depression constant \[={{5.12}^{o}}C\text{ }mo{{l}^{-1}}\] w = weight of solute = 0.44 g \[\Delta {{T}_{f}}={{0.567}^{o}}C\] \[w=22.2g\] \[\therefore \] \[M=\frac{1000\times 5.12\times 0.44}{0.567\times 22.2}=178.9\]


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