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JEE Main & Advanced AIEEE Solved Paper-2008

  • question_answer
    At \[{{80}^{o}}C\], 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 \[{{80}^{o}}C\] and 1 atm pressure, the amount of 'A' in the mixture is (1 atm = 760 mm Hg)       AIEEE  Solved  Paper-2007

    A) 48 mol percent

    B) 50 mol percent

    C) 52 mol percent

    D) 34 mol percent

    Correct Answer: B

    Solution :

     \[{{P}_{T}}=760=P_{A}^{o}{{X}_{A}}+P_{B}^{o}{{X}_{B}}=520{{X}_{A}}+1000\]   \[\,(1-{{X}_{A}})\]            \[{{X}_{A}}=\frac{1}{2}\] or 50 mol percent.


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