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NEET AIPMT SOLVED PAPER SCREENING 2005

  • question_answer
                    A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentarie in the vapour phase would be:                                                                                                                                           

    A)                 0.549               

    B)                 0.200    

    C)                 0.786      

    D)                 0.478

    Correct Answer: D

    Solution :

                    Total vapour pressure of mixture                 = (Mole fraction of pentane V.P. of pentane) + (Mole fraction of hexane x V.P. of hexane) = V.P. of pentane in mixture + V.P. of hexane in mixture                 \[=\left( \frac{1}{5}\times 440+\frac{4}{5}\times 120 \right)=184\,mm\]                         \[\because \,\]            V.P. of pentane in mixture                 = V.P. of mixture x mole fraction of pentane it vapour phase                 88 = 184 x mole fraction of pentane in vapour phase                         \[\therefore \]            Mole fraction of pentane in vapour phase                 \[=\frac{88}{184}=0.478\]


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