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NEET Sample Paper NEET Sample Test Paper-91

  • question_answer
    A solution is prepared by mixing 8.5 g of \[C{{H}_{2}}C{{l}_{2}}\] and 11.95 g of \[CHC{{l}_{3}}.\] If vapour pressure of \[C{{H}_{2}}C{{l}_{2}}\] and \[CHC{{l}_{3}}\]at 298 K are 415 and 200 mmHg respectively, the mole fraction of \[CHC{{l}_{3}}\] in vapour form is: (Molar mass of Cl\[=35.5\,g\,mo{{l}^{-1}}\])

    A) 0.162   

    B)        0.675

    C) 0.325   

    D)        0.486

    Correct Answer: C

    Solution :

    [c] Molar mass of \[CHC{{l}_{3}}=119.5g/mole.\] Molar mass of \[C{{H}_{2}}C{{l}_{2}}=85g/mole.\] Moles of \[CHC{{l}_{3}}=\frac{11.95}{119.5}=0.1\,\,mole.\] Moles of \[C{{H}_{2}}C{{l}_{2}}=\frac{8.5}{85}=0.1\,\,mole.\] Mole fraction of \[CHC{{l}_{3}}=\frac{0.1}{0.2}=0.5\,\,mole.\] Mole fraction of \[C{{H}_{2}}C{{l}_{2}}=\frac{0.1}{0.2}=0.5\,\,mole.\] (Given - Vapour pressure of \[CHC{{l}_{3}}=200\,mm\] Hg = 0.263 atm. Vapour pressure of \[C{{H}_{2}}C{{l}_{2}}=415\,mm\] Hg = 0.546 atm.) (1 atm=760 mm Hg)             \[\therefore \] \[{{P}_{\text{(above}\,\,\text{solution)}}}\] = Mole fraction of \[CHC{{l}_{3}}\times \] (Vapour pressure of \[CHC{{l}_{3}}\]) + Mole fraction of \[C{{H}_{2}}C{{l}_{2}}\times \] (Vapour pressure of \[C{{H}_{2}}C{{l}_{2}}\]) \[=0.5\times 0.263+0.5\times 0.546=0.4045\] Mole fraction of \[CHC{{l}_{3}}\] in vapour form \[=\frac{0.1315}{0.4045}=0.325.\]


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