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

  • question_answer
    1.00 gm of a non-electrolyte solute dissolved in 50 gm of benzene lowered the freezing point of benzene by 0.40 K. \[{{K}_{f}}\] for benzene is 5.12 kg  mol?1. Molecular mass of the solute will be         [DPMT 2004]

    A)                 \[256\,g\,mo{{l}^{-1}}\]   

    B)                 \[2.56\,g\,mo{{l}^{-1}}\]

    C)                 \[512\times {{10}^{3}}\,g\,mo{{l}^{-1}}\] 

    D)                 \[2.56\times {{10}^{4}}\,g\,mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

               By using, \[m=\frac{{{K}_{f}}\times 1000\times w}{\Delta {{T}_{f}}\times {{W}_{\text{Solvent}\,}}(gm)}\]\[=\frac{5.12\,\times 1000\times 1}{0.40\times 50}\]                                \[=256\,gm/mol\]                                 Hence, molecular mass of the solute \[=256\,gm\,mo{{l}^{-1}}\]


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