Question

Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 450 mm Hg at 300 K when mixed in the molar ratio of 1:1 and a vapour pressure of 380 mm Hg when mixed in the molar ratio of 1 : 2 at the same temperature. The vapour pressure of the pure liquid X (in mm Hg) is:

Answer 1

660

Answer 2

The problem involves an ideal solution of two liquids, X and Y, and their total vapor pressures at different molar ratios. We can use Raoult's Law to set up a system of equations.

Raoult's Law: For an ideal solution, the total vapor pressure ($P_T$) is given by: $P_T = P_X^0 \chi_X + P_Y^0 \chi_Y$ where $P_X^0$ and $P_Y^0$ are the vapor pressures of pure liquids X and Y, and $\chi_X$ and $\chi_Y$ are their respective mole fractions.

Given Information:

Case 1: Molar ratio X:Y = 1:1

• Mole fraction of X ($\chi_X$) = $\frac{1}{1+1} = \frac{1}{2}$
• Mole fraction of Y ($\chi_Y$) = $\frac{1}{1+1} = \frac{1}{2}$
• Total vapor pressure ($P_T$) = 450 mm Hg

Applying Raoult's Law: $450 = P_X^0 \left(\frac{1}{2}\right) + P_Y^0 \left(\frac{1}{2}\right)$ Multiplying by 2: $900 = P_X^0 + P_Y^0$ (Equation 1)

Case 2: Molar ratio X:Y = 1:2

• Mole fraction of X ($\chi_X$) = $\frac{1}{1+2} = \frac{1}{3}$
• Mole fraction of Y ($\chi_Y$) = $\frac{2}{1+2} = \frac{2}{3}$
• Total vapor pressure ($P_T$) = 380 mm Hg

Applying Raoult's Law: $380 = P_X^0 \left(\frac{1}{3}\right) + P_Y^0 \left(\frac{2}{3}\right)$ Multiplying by 3: $1140 = P_X^0 + 2P_Y^0$ (Equation 2)

Solving the system of equations: We have two linear equations:

1. $P_X^0 + P_Y^0 = 900$
2. $P_X^0 + 2P_Y^0 = 1140$

Subtract Equation 1 from Equation 2: $(P_X^0 + 2P_Y^0) - (P_X^0 + P_Y^0) = 1140 - 900$ $P_Y^0 = 240$ mm Hg

Substitute the value of $P_Y^0$ into Equation 1: $P_X^0 + 240 = 900$ $P_X^0 = 900 - 240$ $P_X^0 = 660$ mm Hg

The vapor pressure of the pure liquid X is 660 mm Hg.