Methanol (p* = 90 mmHg at 300 K) and ethanol (p* = 51 mmHg at 300 K) form very nearly an ideal solution. The total vapour pressure of a solution obtained by mixing 23 g ethanol with 32 g methanol would be

Methanol (p* = 90 mmHg at 300 K) and ethanol (p* = 51 mmHg at 300 K) form very nearly an ideal solution. The total vapour pressure of a solution obtained by mixing 23 g ethanol with 32 g methanol would be

  1. A

    17 mmHg

  2. B

    60 mmHg

  3. C

    77 mmHg

  4. D

    82 mmHg

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    Solution:

    Amount of methanol, n1=m1M1=32g32 g mol1=1.0mol

    Amount of ethanol, n2=m2M2=23g46 g mol1=0.5 mol

    Mole fraction of methanol, x1=n1n1+n2=1.0mol(1.0+0.5)mol=2/3

    Total Vapour pressure, p=x1p1+x2p2=[(2/3)(90)+(1/3)(51)]mmHg=77mmHg

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