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J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    The vapour pressure of pure liquid solvent is 0.80 arm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.6 arm. What is the mole fraction of component B in this solution?

    A)  0.75             

    B)  0.25

    C)  0.48             

    D)  0.3

    Correct Answer: B

    Solution :

     According to Raoult's law, Relative lowering of vapour pressure \[\propto \]mole fraction of solute or \[\frac{{{p}^{o}}-{{p}_{s}}}{{{p}^{{{o}_{B}}}}}={{\chi }_{B}}\] or \[\frac{0.80-0.60}{0.80}={{\chi }_{B}}\] \[\therefore \] \[{{\chi }_{B}}=\frac{1}{4}=0.25\]


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