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J & K CET Medical J & K - CET Medical Solved Paper-2001

  • question_answer
    An aqueous solution contains 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of acetic acid in this solution:

    A)  0.196                                   

    B)  0.301

    C)  0.392                                   

    D)  0.503

    Correct Answer: B

    Solution :

                    Mole of \[{{C}_{2}}{{H}_{5}}OH=\frac{25}{46}=0.54\] Mole of\[C{{H}_{3}}COOH=\frac{50}{60}=0.83\] % of \[{{H}_{2}}O=100-(50+25)=25\] \[\therefore \]  Mole of \[{{H}_{2}}O=\frac{25}{18}=1.38\] Total number of moles \[=0.54+0.83+1.39\] \[=2.76\] \[\therefore \]Mole fraction of acetic acid \[=\frac{0.83}{2.76}=0.301\]


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