Answer to Question #93884 in Physical Chemistry for Abhishek

Question

Explain the principle of steam distillation. When
a liquid which is immiscible with water was
steam distilled at 95.6°C under a total pressure
of 7.5 x 104Pa, the distillate contained 1.20 g of
the liquid per gram of water. Calculate the
molar mass of the liquid. Vapour pressure of
water is 6.5 x 104Pa at 95-6°C.

Answer 1

Solution.

1)

When a mixture of two practically immiscible liquids is heated, transferring each liquid to the vapor phase, each component creates its own saturated vapor pressure as if the other component were absent. Consequently, the vapor pressure of the entire system increases. Boiling begins when the sum of the vapor pressures of two immiscible liquids simply exceeds atmospheric pressure. This is the principle of steam distillation.

2)

$P = P(H2O) + P(x)$

$P(x) = P - P(H2O)$

$P(x) = 1 \times 10^4$

X1 is mole fraction of unknown component.

$X1 = \frac{P(x)}{P}$

X2 is mole fraction of water.

$X2 = 1 - X1$

X2 = 0.87

$X2 = \frac{n(H2O)}{n(x) + n(H2O)}$

$X2 = \frac{\frac{1}{18}}{\frac{1.20}{M} + \frac{1}{18}}$

M = 144.55 g/mole

Answer:

1)

When a mixture of two practically immiscible liquids is heated, transferring each liquid to the vapor phase, each component creates its own saturated vapor pressure as if the other component were absent. Consequently, the vapor pressure of the entire system increases. Boiling begins when the sum of the vapor pressures of two immiscible liquids simply exceeds atmospheric pressure. This is the principle of steam distillation.

2)

M = 144.55 g/mole