Question

The correct option for the value of vapor pressure of a solution at 45\(^{\circ}\)C with benzene to octane in molar ratio 3 : 2 is : [At 45\(^{\circ}\)C vapor pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

• A. 350 mm of Hg
• B. 160 mm of Hg
• C. 168 mm of Hg
• D. 336 mm of Hg

Answer 1

The Correct Option is D

To find the vapor pressure of the solution at 45°C with a molar ratio of benzene to octane of 3:2, we can use Raoult’s Law. Raoult’s Law states that the vapor pressure of a solution is directly related to the vapor pressures of the pure components and their mole fractions in the solution.

Raoult's Law formula is given as:

\(P_{\text{solution}} = x_1 P_1^0 + x_2 P_2^0\) 

where:

• \( P_{\text{solution}} \) is the vapor pressure of the solution.
• \( x_1 \) and \( x_2 \) are the mole fractions of benzene and octane respectively.
• \( P_1^0 \) and \( P_2^0 \) are the vapor pressures of pure benzene and octane, respectively.

Given:

• Vapor pressure of pure benzene, \( P_1^0 = 280 \) mm Hg.
• Vapor pressure of pure octane, \( P_2^0 = 420 \) mm Hg.
• Mole ratio of benzene to octane is 3:2.

First, calculate the total number of moles:

\(3 + 2 = 5\text{ moles}\)

Next, calculate the mole fractions:

• \( x_1 = \frac{3}{5} = 0.6 \) (mole fraction of benzene)
• \( x_2 = \frac{2}{5} = 0.4 \) (mole fraction of octane)

Now, apply Raoult's Law to calculate the vapor pressure of the solution:

\(P_{\text{solution}} = (0.6 \times 280) + (0.4 \times 420)\)

Calculate each term separately:

• \( 0.6 \times 280 = 168 \) mm Hg
• \( 0.4 \times 420 = 168 \) mm Hg

Add both contributions to find the total vapor pressure of the solution:

\(P_{\text{solution}} = 168 + 168 = 336 \) mm\)

Therefore, the correct option for the vapor pressure of the solution at 45°C is 336 mm of Hg.