Question

The vapour pressure of pure liquids A and B are 450 and 700 mm of Hg at 350 K respectively. If the total vapour pressure of the mixture is 600 mm of Hg, the composition of the mixture in the solution is

• A. x_A = 0.4, x_B = 0.6
• B. x_A = 0.6, x_B = 0.4
• C. x_A = 0.3, x_B = 0.7
• D. x_A = 0.7, x_B = 0.3

Answer 1

The Correct Option is A

According to Raoult’s Law, the total vapor pressure of a mixture is given by: 

P_total = P_A + P_B

Where:

• P_A = x_A * P_A0
• P_B = x_B * P_B0

We are given that:

• P_A0 = 450 mm Hg (vapor pressure of pure A)
• P_B0 = 700 mm Hg (vapor pressure of pure B)
• P_total = 600 mm Hg

Substituting the values into Raoult’s law:

600 = x_A * 450 + x_B * 700

Since x_A + x_B = 1, we have:

600 = x_A * 450 + (1 - x_A) * 700

Expanding and solving for x_A:

600 = 450 * x_A + 700 - 700 * x_A

600 = 700 - 250 * x_A

250 * x_A = 100

x_A = 0.4

Therefore, x_B = 1 - x_A = 0.6

The correct answer is (A) : x_A = 0.4, x_B = 0.6.

Answer 2

1. Recall Raoult's Law
Raoult's Law states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

$P_A = x_A P_A^0$
$P_B = x_B P_B^0$

where:
- $P_A$ and $P_B$ are the partial vapor pressures of components A and B, respectively.
- $x_A$ and $x_B$ are the mole fractions of components A and B in the liquid mixture, respectively.
- $P_A^0$ and $P_B^0$ are the vapor pressures of pure components A and B, respectively.

2. Apply Raoult's Law to the mixture
The total vapor pressure of the mixture is the sum of the partial vapor pressures of the components:

$P_{total} = P_A + P_B = x_A P_A^0 + x_B P_B^0$

We also know that the sum of the mole fractions is equal to 1:

$x_A + x_B = 1$, so $x_B = 1 - x_A$

3. Substitute the given values
We are given:
- $P_A^0 = 450$ mm Hg
- $P_B^0 = 700$ mm Hg
- $P_{total} = 600$ mm Hg

Substitute $x_B = 1 - x_A$ into the total pressure equation:

$P_{total} = x_A P_A^0 + (1 - x_A) P_B^0$
$600 = x_A (450) + (1 - x_A) (700)$

4. Solve for $x_A$
$600 = 450x_A + 700 - 700x_A$
$600 - 700 = 450x_A - 700x_A$
$-100 = -250x_A$
$x_A = \frac{-100}{-250} = \frac{100}{250} = \frac{2}{5} = 0.4$

5. Solve for $x_B$
$x_B = 1 - x_A = 1 - 0.4 = 0.6$

6. Determine the composition
The composition of the mixture is $x_A = 0.4$ and $x_B = 0.6$.

7. Compare with the given options
Comparing the derived composition with the options, we find that it matches option (A).

Final Answer:
(A) $x_A = 0.4, x_B = 0.6$