JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    A solution at \[20{}^\circ C\] is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively : JEE Main Online Paper (Held On 10 April 2015)

    A)  30.5 torr and 0.389

    B)  38.0 torr and 0.589

    C)  35.8 torr and 0.280

    D)  4 35.0 torr and 0.480

    Correct Answer: C

    Solution :

    Total V.P. of solution \[=P_{A}^{0}{{X}_{A}}+P_{B}^{0}{{X}_{B}}\] Total V.P. of solution \[=\left( \frac{1.5}{5}\times 74.7+\frac{3.5}{5}\times 22.3 \right)torr\] \[=\left( 15.6+22.4 \right)torr\,=38torr\] Mole fraction of Benzene in vapour form \[=\frac{22.4}{38}=0.589\]


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