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JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    The vapour pressure of two liquids \[P\] and \[Q\] are \[80\] and \[60\]\[Torr\], respectively. The total vapour pressure of solution obtained by mixing \[3\] moles of \[P\] and \[2\] moles of \[Q\] would be

    A) \[140\,\,Torr\]

    B) \[20\,\,Torr\]

    C) \[68\,\,Torr\]

    D)  \[72\,\,Torr\]

    Correct Answer: D

    Solution :

    Mole fraction of\[P=\frac{3}{3+2}=\frac{3}{5}\] Mole fraction of\[Q=\frac{2}{3+2}=\frac{2}{5}\] Hence, total vapour pressure = Mole fraction of\[P\times V.P\]\[\text{of}\]\[P\]                 + mole fraction of\[Q\times V.P.\]\[\text{of}\]\[Q\]                 \[=\left( \frac{3}{5}\times 80+\frac{2}{5}\times 60 \right)=48+24\]                 \[=72\,\,Torr\]


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