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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    The freezing point of a solution, prepared from \[1.25\text{ }gm\]. of a non-electrolyte and \[20gm\] of water, is \[271.9\text{ }K\]. If the molar depression constant is \[1.86\text{ }K/mol,\] then molar mass of the solute is

    A)  \[103.3\]                            

    B)         \[105.7\]

    C)         \[107.9\]                            

    D)         \[109.3\]

    Correct Answer: B

    Solution :

    The molar mass of solute \[=\frac{1000\times {{K}_{f}}\times W}{\Delta {{T}_{f}}\times \omega }\] Here   weight of solute \[(W)=1.25g\] weight of solvent \[(\omega )=20g\] Molar depression constant                 \[({{K}_{f}})=1.86\] Freezing point of a solution                 \[(\Delta {{T}_{f}})=(273-271.9)\]                 \[=1.1K\] Putting the values, the molar mass of the solute                 \[=\frac{1000\times 1.86\times 1.25}{1.1\times 20}=105.7\]


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