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 determine the vapor pressure of the solution at 45\(^{\circ}\)C, we will use Raoult's law. Raoult's law states that the vapor pressure of an ideal solution is dependent on the vapor pressures of each component and their mole fractions in the solution.

The formula for the total vapor pressure of a solution is given by:

\(P_{\text{solution}} = X_{\text{benzene}} \cdot P_{\text{benzene}}^0 + X_{\text{octane}} \cdot P_{\text{octane}}^0\)

where:

• \(X_{\text{benzene}}\) and \(X_{\text{octane}}\) are the mole fractions of benzene and octane, respectively.
• \(P_{\text{benzene}}^0 = 280 \text{ mm Hg}\)
• \(P_{\text{octane}}^0 = 420 \text{ mm Hg}\)

Given that the molar ratio of benzene to octane is 3:2, the total number of moles is \(3 + 2 = 5\).

Therefore, the mole fractions are:

• \(X_{\text{benzene}} = \frac{3}{5}\)
• \(X_{\text{octane}} = \frac{2}{5}\)

Substitute these into the equation for \(P_{\text{solution}}\):

\(P_{\text{solution}} = \left(\frac{3}{5}\right) \cdot 280 + \left(\frac{2}{5}\right) \cdot 420\)

Calculate each term:

\(\left(\frac{3}{5}\right) \cdot 280 = 168\)

\(\left(\frac{2}{5}\right) \cdot 420 = 168\)

Therefore, the total vapor pressure is:

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

Thus, the correct answer is 336 mm of Hg.