Question

At T(K) vapour pressure of pure benzene and toluene are 500 and 200 mm Hg respectively. If they form an ideal solution, what is the mole fraction of toluene in a mixture boiling at T(K) at a total pressure of 380 mm Hg ?

• A. \(0.20 \)
• B. \(0.60 \)
• C. \(0.40 \)
• D. \(0.80 \)

Hint:

For ideal solutions, remember Raoult's Law: \( P_{\text{total}} = \sum (x_i P^0_i) \). This law is fundamental for solving vapor pressure problems involving ideal binary or multi-component mixtures.

Answer 1

The Correct Option is C

Step 1: Identify the given information.
We are given the following values:
Vapour pressure of pure benzene (\( P^0_{\text{benzene}} \)) = 500 mm Hg
Vapour pressure of pure toluene (\( P^0_{\text{toluene}} \)) = 200 mm Hg
Total vapor pressure of the solution (\( P_{\text{total}} \)) = 380 mm Hg
The solution is ideal.
We need to find the mole fraction of toluene in the liquid phase (\( x_{\text{toluene}} \)). Step 2: Apply Raoult's Law for ideal solutions.
For an ideal solution, Raoult's Law states that the partial vapor pressure of each component is equal to the product of its mole fraction in the liquid phase and its vapor pressure in the pure state. \[ P_{\text{benzene}} = x_{\text{benzene}} \times P^0_{\text{benzene}} \] \[ P_{\text{toluene}} = x_{\text{toluene}} \times P^0_{\text{toluene}} \] The total vapor pressure of the solution is the sum of the partial vapor pressures: \[ P_{\text{total}} = P_{\text{benzene}} + P_{\text{toluene}} \] Substitute the expressions for partial pressures: \[ P_{\text{total}} = (x_{\text{benzene}} \times P^0_{\text{benzene}}) + (x_{\text{toluene}} \times P^0_{\text{toluene}}) \] Step 3: Relate mole fractions in the liquid phase.
For a binary solution of benzene and toluene, the sum of their mole fractions in the liquid phase is 1: \[ x_{\text{benzene}} + x_{\text{toluene}} = 1 \] So, we can write \( x_{\text{benzene}} = 1 - x_{\text{toluene}} \). Step 4: Substitute values into the total vapor pressure equation and solve for \( x_{\text{toluene}} \).
Substitute \( x_{\text{benzene}} = 1 - x_{\text{toluene}} \) into the total vapor pressure equation: \[ 380 = ((1 - x_{\text{toluene}}) \times 500) + (x_{\text{toluene}} \times 200) \] \[ 380 = 500 - 500x_{\text{toluene}} + 200x_{\text{toluene}} \] Combine the terms with \( x_{\text{toluene}} \): \[ 380 = 500 - 300x_{\text{toluene}} \] Rearrange the equation to solve for \( x_{\text{toluene}} \): \[ 300x_{\text{toluene}} = 500 - 380 \] \[ 300x_{\text{toluene}} = 120 \] \[ x_{\text{toluene}} = \frac{120}{300} \] \[ x_{\text{toluene}} = \frac{12}{30} = \frac{2}{5} = 0.4 \] Thus, the mole fraction of toluene in the mixture is 0.40. Step 5: Verify the answer with the options.
The calculated mole fraction of toluene is 0.40, which matches option (3).