Raoult S Law And Henry S Law

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.