Question

An aqueous solution which contains 42% by weight (w/W) of a volatile liquid "A" of molar mass of 140 g/mol, has a vapour pressure of 2141.4 mm of Hg at 37°C. What is the vapour pressure of the pure liquid "A"? Vapour pressure of water at 37°C is 2019.1 mm.

• A. 3457.6 mm
• B. 293 mm
• C. 2356.4 mm
• D. 2519.2 mm

Hint:

To calculate the vapour pressure of a pure component in a solution, use Raoult's law. This law relates the vapour pressure of a solution to the mole fraction of the solute and the vapour pressures of the pure components.

Answer 1

The Correct Option is A

Step 1: Use Raoult's Law.
Raoult's law states that the vapour pressure of a solution is the sum of the vapour pressures of the individual components. The equation is:
\[ P_{\text{solution}} = x_A \cdot P_A^{\text{pure}} + x_B \cdot P_B^{\text{pure}} \]
Where:
- \( P_{\text{solution}} \) is the total vapour pressure of the solution.
- \( x_A \) and \( x_B \) are the mole fractions of the components in the solution.
- \( P_A^{\text{pure}} \) and \( P_B^{\text{pure}} \) are the vapour pressures of the pure components.
In our case, the solution contains liquid "A" and water. The vapour pressure of the solution is given as 2141.4 mm of Hg.

Step 2: Calculate the mole fraction of liquid A.
We are given: - The mass fraction of "A" is 42% by weight, i.e., \( w_A = 0.42 \).
- The molar mass of "A" is 140 g/mol, and the molar mass of water is 18 g/mol.
Let the total mass of the solution be 100 g. Thus:
- The mass of "A" is \( 0.42 \times 100 = 42 \) g.
- The mass of water is \( 100 - 42 = 58 \) g.
The number of moles of "A" is:
\[ n_A = \frac{42}{140} = 0.3 \, \text{mol} \]
The number of moles of water is: \[ n_B = \frac{58}{18} = 3.22 \, \text{mol} \]
The total number of moles in the solution is:
\[ n_{\text{total}} = n_A + n_B = 0.3 + 3.22 = 3.52 \, \text{mol} \]
The mole fraction of "A" is: \[ x_A = \frac{n_A}{n_{\text{total}}} = \frac{0.3}{3.52} = 0.0852 \]

Step 3: Use Raoult's Law to find the vapour pressure of pure "A".
According to Raoult's law, the vapour pressure of the solution is given by:
\[ P_{\text{solution}} = x_A \cdot P_A^{\text{pure}} + x_B \cdot P_B^{\text{pure}} \]
Substitute the known values into the equation: \[ 2141.4 = 0.0852 \cdot P_A^{\text{pure}} + (1 - 0.0852) \cdot 2019.1 \]
Solving for \( P_A^{\text{pure}} \): \[ 2141.4 = 0.0852 \cdot P_A^{\text{pure}} + 0.9148 \cdot 2019.1 \]
\[ 2141.4 = 0.0852 \cdot P_A^{\text{pure}} + 1843.76 \]
\[ 2141.4 - 1843.76 = 0.0852 \cdot P_A^{\text{pure}} \]
\[ 297.64 = 0.0852 \cdot P_A^{\text{pure}} \] \[ P_A^{\text{pure}} = \frac{297.64}{0.0852} = 3457.6 \, \text{mm} \]

Step 4: Conclusion.
Thus, the vapour pressure of the pure liquid "A" is \( 3457.6 \, \text{mm} \), which corresponds to option (A).