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MGIMS WARDHA MGIMS WARDHA Solved Paper-2013

  • question_answer
    60 g of urea is dissolved in 1100 g solution. To keep\[\Delta {{T}_{f}}/{{K}_{f}}\]as 1 mol\[k{{g}^{-1}}\], water separated in the form of ice is

    A)  40 g                                      

    B)  60 g

    C)  100 g                                    

    D)  200 g

    Correct Answer: A

    Solution :

                    1100 g solution has 60 g urea Water =1040 g                 \[\Delta {{T}_{f}}=\frac{1000{{K}_{f}}{{w}_{2}}}{{{M}_{2}}{{w}_{1}}}\]                 \[\frac{\Delta {{T}_{f}}}{{{K}_{f}}}=\frac{1000\times 60}{60\times {{w}_{1}}}=1\] or            \[{{w}_{1}}=1000\,g\] The ice formed \[=1040-1000=40\text{ }g\]


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