JEE Main & Advanced Sample Paper JEE Main Sample Paper-33

  • question_answer
    A mixture of [a] and [b], which are two miscible liquids, is under equilibrium conditions at atmospheric pressure. The mole fraction of [a] in solution is 0.3 and in vapour phase is 0.6. If the solution behaves ideally the ratio of \[P_{a}^{o}\] to \[P_{b}^{o}\] is:

    A)  1.5   

    B)                                     3.5

    C)  2.5         

    D)                     4.0

    Correct Answer: B

    Solution :

                In solution \[{{X}_{a}}=0.3\]            \[\therefore \,\,\,\,\,\,\,\,{{X}_{b}}=0.7\]             In vapors \[{{Y}_{a}}=0.6\] \[\therefore \,\,\,\,\,\,{{Y}_{b}}=0.4\]             Using Dalton?s law             \[{{Y}_{a}}\,=\frac{{{P}_{a}}}{{{P}_{T}}}\]             i.e.,      \[{{Y}_{a}}\,=\frac{P_{a}^{o}\times {{X}_{a}}}{P_{a}^{o}\times \,{{X}_{a}}\,\times P_{b}^{o}\,{{X}_{b}}}\]             \[\therefore \,\,\,{{Y}_{b}}\,=\frac{P_{b}^{o}\,\times {{X}_{b}}}{P_{a}^{o}\times \,{{X}_{a}}\,\times P_{b}^{o}\,{{X}_{b}}}\]             \[\therefore \,\,\,\frac{{{Y}_{a}}}{{{Y}_{b}}}\,=\frac{P_{a}^{o}{{X}_{a}}}{P_{b}^{o}\,{{X}_{b}}}\]             \[\therefore \,\,\frac{P_{a}^{o}}{P_{b}^{o}\,}=\frac{{{Y}_{a}}}{{{Y}_{b}}}\times \frac{{{X}_{b}}}{{{X}_{a}}}\,=\frac{0.6}{0.4}\,\times \frac{0.7}{0.4}\times \,\frac{0.7}{\,0.3}\,=3.5\]


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