Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is: (xM = Mole fraction of 'M' in solution ; xN = Mole fraction of 'N' in solution ; yM = Mole fraction of 'M' in vapour phase ; yN = Mole fraction of 'N' in vapour phase)

Engineering

Chemistry

Ideal Solution

Raoult Law and Vapour Pressure

Question

Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is:

(x_M = Mole fraction of 'M' in solution ;

x_N = Mole fraction of 'N' in solution ;

y_M = Mole fraction of 'M' in vapour phase ;

y_N = Mole fraction of 'N' in vapour phase)

$\frac{x_{M}}{x_{N}} = \frac{y_{M}}{y_{N}}$

$\frac{x_{M}}{x_{N}} > \frac{y_{M}}{y_{N}}$

$\frac{x_{M}}{x_{N}} < \frac{y_{M}}{y_{N}}$

(x_M – y_M) < (x_N – y_N)

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$P_{M}^{0} = 450$

$P_{N}^{0} = 700$

P_M = x_M 450

P_N = x_N 700

Y_MP_T = P_M

Y_NP_T = P_N

_{$\frac{Y_{M} P_{T}}{Y_{N} P_{T}} = \frac{x_{M}}{x_{N}} \frac{450}{700}$}

_{$\frac{Y_{M}}{Y_{N}} = \frac{x_{M}}{x_{N}} (0 : 64)$}

_{$\frac{Y_{M}}{Y_{N}} > \frac{x_{M}}{x_{N}}$}