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

  • question_answer
    At \[80{}^\circ C\], the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture solution of 'A' and 'B' boils at \[80{}^\circ C\] and 1 atm pressure, the amount of 'A' in the mixture is (1 atm = 760 mm Hg)

    A) 52 mol percent              

    B) 34 mol percent

    C) 48 mol percent              

    D) 50 mol percent

    Correct Answer: D

    Solution :

    [d] At 1 atmospheric pressure the boiling point of mixture is \[80{}^\circ C.\] At boiling point the vapour pressure of mixture, \[{{P}_{T}}=1\] atmosphere =760 mm Hg. Using the relation, \[{{P}_{T}}=P_{A}^{{}^\circ }{{x}_{A}}+P_{B}^{{}^\circ }{{x}_{B}}\], we get \[{{P}_{T}},=520{{x}_{A}}+1000(1-{{x}_{A}})\] \[\{\therefore P_{A}^{0}=520\,mm\,Hg,\] \[P_{B}^{0}=1000\text{ }mm\text{ }Hg;{{x}_{A}}+{{x}_{B}}=1\] or \[760=520{{x}_{A}}+1000-1000{{x}_{A}}\,or\,480{{x}_{A}}=240\] or\[{{X}_{A}}=\frac{240}{480}=\frac{1}{2}\] or  50 mol percent


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