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NEET Chemistry Solutions / विलयन Question Bank Self Evaluation Test - Solutions

  • question_answer
    A solution of urea (mol. mass\[56g\text{ }mo{{l}^{-1}}\]) boils at \[100.18{}^\circ C\] at the atmospheric pressure. If \[{{K}_{f}}\] and \[{{K}_{b}}\] for water are 1.86 and \[0.512K\text{ }kg\text{ }mo{{l}^{-1}}\] respectively, the above solution will freeze at

    A) \[0.654\text{ }{}^\circ C\]        

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

    C) \[6.54\text{ }{}^\circ C\]         

    D) \[-6.54{}^\circ C\]

    Correct Answer: B

    Solution :

    [b] As \[\Delta {{T}_{f}}={{K}_{f}}m\] \[\Delta {{T}_{b}}={{K}_{b}}.m\] Hence, we have \[m=\frac{\Delta {{T}_{f}}}{{{K}_{f}}}=\frac{\Delta {{T}_{b}}}{{{K}_{b}}}\] or \[\Delta {{T}_{f}}=\Delta {{T}_{b}}\frac{{{K}_{f}}}{{{K}_{b}}}\] \[\left[ \Delta {{T}_{b}}=100.18-100=0.18{}^\circ C \right]\] \[=0.18\times \frac{1.86}{0.512}=0.654{}^\circ C\] 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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