JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • question_answer
    A solution of urea \[\left( mol.\text{ }mass\text{ }56\text{ }g\text{ }mo{{l}^{-}}^{1} \right)\] boils at 100. \[18{}^\circ C\] at the atmospheric pressure. If \[{{K}_{f}}\,\,and\,\,{{K}_{b}}\] for water are 1.86 and 0.512 K kg \[mo{{l}^{-1}}\] respectively, the above solution will freeze at

    A) \[0.654\,{}^\circ C\]     

    B)        \[-\,0.654\,{}^\circ C\]

    C) \[\,6.54\,{}^\circ C\]     

    D)        \[-\,6.54\,{}^\circ C\]

    Correct Answer: B

    Solution :

    As \[\Delta {{T}_{f}}={{K}_{f}}\,m\] \[\Delta {{T}_{b}}=\,\,{{K}_{b}}.\,m\] Hence, we have \[m=\frac{\Delta {{T}_{f}}}{{{K}_{f}}}\,\,=\,\,\frac{\Delta {{T}_{b}}}{{{K}_{b}}}\] \[\Delta {{T}_{f}}=\Delta {{T}_{b}}\frac{{{K}_{f}}}{{{K}_{b}}}\] \[\Rightarrow \,\,\,[\Delta {{T}_{b}}=100.18-100=0.18{}^\circ \,C]\] \[=\,\,0.18\times \frac{1.86}{0.512}=0.654{}^\circ \,C\] As the Freezing Point of pure water is \[0{}^\circ C\], \[\Delta {{T}_{f}}=0-{{T}_{f}}\] \[0.654=\,\,0-{{T}_{f}}.\] \[\therefore \,\,{{T}_{f}}=-\,0.654\] Thus the freezing point of solution will be \[-\,0.654{}^\circ C\].

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