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J & K CET Medical J & K - CET Medical Solved Paper-2010

  • question_answer
    The vapour pressure of pure liquid solvent is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.6 atm. 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 Raoults law, Relative lowering of vapour pressure \[\propto \]mole fraction of solute Or                           \[\frac{{{p}^{o}}-{{p}_{s}}}{{{p}^{o}}}={{\chi }_{B}}\] Or                           \[\frac{0.80-0.60}{0.80}={{\chi }_{B}}\] \[\therefore \]                  \[{{\chi }_{B}}=\frac{1}{4}=0.25\]


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