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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    The relative lowering of vapour pressure of a dilute aqueous solution containing nonvolatile solute is 0.0125. The molality of the solution is about

    A) 0.70                                       

    B) 0.50

    C) 0.90                       

    D)        0.80

    Correct Answer: A

    Solution :

    Relative lowering of vapour pressure = mole fraction of solute (Raoults law)                 \[\frac{p-{{p}_{s}}}{p}={{\chi }_{2}}\]                 \[\frac{p-{{p}_{s}}}{p}=\frac{wM}{mW}\]                 \[0.0125=\frac{wM}{mW}\] or            \[\frac{w}{mW}=\frac{0.0125}{18}=0.00070\] Hence,  \[molality=\frac{w}{mW}\times 1000\]                 \[=0.0007\times 1000=0.70\]


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