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NEET Chemistry Solutions / विलयन Question Bank Self Evaluation Test - Solutions

  • question_answer
    The vapour pressure of two pure liquids A and B that form an ideal solution, are 400 and 800 mm of Hg respectively at a temperature \[t{}^\circ C.\] The mole fraction of A in a solution of A and B whose boiling point is \[t{}^\circ C.\] will be

    A) 0.4                   

    B) 0.8  

    C) 0.1                   

    D) 0.2

    Correct Answer: C

    Solution :

    [c] V.P. of solution at \[t{}^\circ C=760\text{ }mm\] [at b.p., V.P. of solution =atompheric pressure] Thus \[=P_{A}^{{}^\circ }.{{x}_{A}}+P_{B}^{{}^\circ }.{{x}_{B}}\] or \[P=P_{A}^{{}^\circ }.{{X}_{A}}+P_{B}^{{}^\circ }.(1-{{x}_{A}})\,[\therefore {{x}_{A}}+{{x}_{B}}=1]\] or \[760=400{{X}_{A}}+800(1-{{X}_{A}})\]                                     \[\left[ \therefore P=760\text{ }mm\,of\,Hg \right]\] or \[-800+760=-400{{x}_{A}}\] or \[-40=-400{{x}_{A}}\] or \[{{x}_{A}}=\frac{40}{400}=0.1\] Thus mole fraction of in solution is 0.1


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