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NEET Sample Paper NEET Sample Test Paper-73

  • question_answer
    The vapour pressures of ethanol and methanol are 42.0 mm and 88.5 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol. The mole fraction of methanol in the vapour is:

    A) 0.467               

    B) 0.502

    C) 0.513               

    D) 0.556

    Correct Answer: C

    Solution :

    \[{{P}_{m}}=P_{MeOH}^{0}X_{MeOH}^{0}+P_{EtOH}^{0}X_{EtOH}^{{}}\] \[{{P}_{m}}=88.5\times \left[ \frac{\frac{16}{32}}{\frac{16}{32}+\frac{46}{46}} \right]+42X;\,\,\left[ \frac{\frac{46}{46}}{\frac{16}{32}+\frac{46}{46}} \right]\,\,=\,\,57.5\]Now \[P{{'}_{MeOH}}=={{P}_{m}}\cdot {{X}_{MeOH(g)}}\] \[\therefore \,\,\,\,\,88.5\,\times =57.5\,\,{{X}_{MeOH(g)}}\] \[K=\left[ \frac{\frac{16}{32}}{\frac{16}{32}+\frac{46}{46}} \right]\] \[\therefore \,\,\,\,\,\,\,\,{{X}_{MeOH}}\,\,=\,\,0.513\]


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