Question

At 20°C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20°C above an equimolar mixture of benzene and methylbenzene is ______ × 10^-2. (Nearest integer)

विकल्प

• 52

• 96

• 78

• 65

Answer 1

At 20°C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20°C above an equimolar mixture of benzene and methylbenzene is 78 × 10^-2.

Explanation:

Given:

`"P"_"B"^"o"` = 70 torr

`"P"_"M"^"o"` = 20 torr

The mole fraction of benzene and methylbenzene in the equimolar solution of these two chemicals is 0.5 for each.

x_B = x_M = 0.5

Suppose that a mixture of benzene and methylbenzene at an equimolar ratio has a mole fraction of y_B in the vapour phase at 20°C.

The following formula can be used to calculate the mole fraction of benzene:

`"y"_"B" = ("x"_"B""P"_"B"^"o")/("x"_"B""P"_"B"^"o" + "x"_"M""P"_"M"^"o")`

Here, x_B and x_M denote the mole fractions of benzene and methylbenzene respectively. The `"P"_"B"^"o"` and `"P"_"M"^"o"` denotes the vapour pressure of benzene and methylbenzene respectively. Substituting respective values in the above formula, we get,

`"y"_"B" = (0.5 xx 70)/(0.5 xx 70 + 0.5 xx 20)`

`therefore "y"_"B"` = 0.777 = 77.7 × 10^-2 

`therefore "y"_"B"` = 78 × 10^-2