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

  • question_answer
    The freezing point of a solution prepared from \[1.25\,gm\] of a non-electrolyte and \[20\,gm\] of water is \[271.9\,K\]. If molar depression constant is \[1.86K\,mol{{e}^{-1}}\], then molar mass of the solute will be              [AFMC 1998; CPMT 1999]

    A)                 105.7     

    B)                 106.7

    C)                 115.3     

    D)                 93.9

    Correct Answer: A

    Solution :

             Molar mass \[=\frac{{{K}_{f}}\times 1000\times w}{\Delta {{T}_{f}}\times W}=\frac{1.86\times 1000\times 1.25}{20\times 1.1}\]                                 \[=105.68=105.7\].            


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