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J & K CET Medical J & K - CET Medical Solved Paper-2009

  • question_answer
    At 300 K, two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is

    A)  0.6                                        

    B)  0.5

    C)  0.8                                        

    D)  0.4

    Correct Answer: D

    Solution :

                    Let the moles of\[A=x\] \[\therefore \]The moles of\[B=x\][\[\because \]mixture is equimolar.] Mole fraction of \[A=\frac{x}{x+x}=0.5=\] mole fraction of B. Thus,\[{{P}_{T}}=p_{A}^{o}{{X}_{A}}+p_{B}^{o}{{X}_{B}}\] \[{{p}_{T}}=150\times 0.5+100\times 0.5\] \[=125\] \[\therefore \]Mole fraction of B in vapour mixture                 \[=\frac{p_{B}^{o}{{X}_{B}}}{{{p}_{T}}}\]                 \[=\frac{100\times 0.5}{125}\] \[=0.4\]


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