BHU PMT BHU PMT (Screening) Solved Paper-2011

  • question_answer
    18 g of glucose is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at\[100{}^\circ C\]is

    A)  759.2 torr           

    B)  760.5 torr

    C)  76.8 torr             

    D)  752.40 torr

    Correct Answer: D

    Solution :

                     \[\frac{{{p}^{o}}-{{p}_{s}}}{{{p}^{o}}}=\frac{n}{N}\] \[\frac{760-{{p}_{s}}}{760}=\frac{\frac{18}{180}}{\frac{178.2}{18}}=\frac{0.1}{9.9}\] \[760-{{p}_{s}}=\frac{1}{99}\times 760\] \[{{p}_{s}}=760-7.6=752.4\]


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