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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Lowering of vapour pressure

  • question_answer
    The vapour pressure of pure liquid A is 0.80 atm. On mixing a non-volatile B to A, its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is        [MP PET 2003]

    A)                 0.150     

    B)             0.25

    C)                 0.50       

    D)                 0.75

    Correct Answer: B

    Solution :

             According to Raoult?s Law            \[\frac{{{P}^{0}}-{{P}_{s}}}{{{P}^{0}}}={{x}_{B}}\]  (Mole fraction of solute)                 \[{{x}_{B}}=\frac{0.8-0.6}{0.8}=0.25\].


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