The vapour pressure of pure liquids A and B are 450 and 700 mm Hg res…

The correct answer is option 1. X_A = 0.4, X_B = 0.6.

To find the composition of the liquid mixture (mole fractions of A and B) when the total vapor pressure is 600 mm Hg, we can use Dalton's Law of partial pressures. According to Dalton's Law, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.

Let us denote:

\( P_A \) as the partial pressure of component A,

\( P_B \) as the partial pressure of component B,

\( x_A \) as the mole fraction of component A,

\( x_B \) as the mole fraction of component B.

Given:

Vapor pressure of pure liquid A (\( P^{\circ}_A \)) = 450 mm Hg

Vapor pressure of pure liquid B (\( P^{\circ}_B \)) = 700 mm Hg

Total vapor pressure of the mixture (\( P_{\text{total}} \)) = 600 mm Hg

We can use the equation:

\(P_{\text{total}} = P_A + P_B \)

Using Raoult's Law, we have:

\(P_A = x_A \cdot P^{\circ}_A \)

\( P_B = x_B \cdot P^{\circ}_B \)

Substituting these equations into the total pressure equation:

\(P_{\text{total}} = x_A \cdot P^{\circ}_A + x_B \cdot P^{\circ}_B \)

Given that \( P_{\text{total}} = 600 \) mm Hg, \( P^{\circ}_A = 450 \) mm Hg, and \( P^{\circ}_B = 700 \) mm Hg, we can solve for \( x_A \) and \( x_B \).

\(600 = x_A \cdot 450 + x_B \cdot 700 \)

Now, we can rearrange this equation to solve for \( x_B \):

\(600 - 450x_A = 700x_B \)

\( x_B = \frac{600 - 450x_A}{700} \)

Since \( x_A + x_B = 1 \), we can substitute \( x_B \) from the above equation into \( x_A + x_B = 1 \) and solve for \( x_A \):

\(x_A + \frac{600 - 450x_A}{700} = 1 \)

\(700x_A + 600 - 450x_A = 700 \)

\(250x_A = 100 \)

\( x_A = \frac{100}{250} = 0.4 \)

Now, we can find \( x_B \):

\(x_B = 1 - x_A = 1 - 0.4 = 0.6 \)

Therefore, the composition of the liquid mixture is \( x_A = 0.4 \) and \( x_B = 0.6 \).

So, the correct option is \(X_A = 0.4, X_B = 0.6\)