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NEET Sample Paper NEET Sample Test Paper-90

  • question_answer
    Vapour pressure (in ton-) of an ideal solution of two liquids A and B is given by: \[P=52{{X}_{A}}+114\] where \[{{X}_{A}}\] is the mole fraction of A in the mixture. The vapour pressure (in torr) of equimolar mixture of the two liquids will be:

    A) 166                              

    B) 83    

    C) 140      

    D)         280

    Correct Answer: C

    Solution :

    [c] Total V.P., \[P=P_{A}^{{}^\circ }{{X}_{A}}+P_{B}^{{}^\circ }{{X}_{B}}=P_{A}^{{}^\circ }{{X}_{A}}+P_{B}^{{}^\circ }(1-{{X}_{A}})\]     \[=(P_{A}^{{}^\circ }-P_{B}^{{}^\circ }){{X}_{A}}+P_{B}^{{}^\circ }\] Thus, \[P_{B}^{{}^\circ }=114\,\,torr;\] \[P_{A}^{{}^\circ }-P_{B}^{{}^\circ }=52\] or  \[P_{A}^{{}^\circ }=166\,\,torr\] Hence \[P=166\times \frac{1}{2}+114\times \frac{1}{2}=140\,\,torr\]


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