An aqueous solution containing liquid `A(M.Wt.=128) 64%` by weight has a vapour pressure of `145mm`. Find the vapour pressure A. If that of water is `155mm` at the same temperature.

Explanation

To find the vapor pressure of liquid A, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution, plus the vapor pressure of the solute multiplied by its mole fraction in the solution.

### Step-by-step Solution:

1. **Identify Given Values:**
- Vapor pressure of the solution, \( P_{\text{total}} = 145 \, \text{mmHg} \)
- Vapor pressure of pure water, \( P^0_{\text{H}_2\text{O}} = 155 \, \text{mmHg} \)
- Mass percent of solute A = 64%
- Molecular weight of A = 128 g/mol

2. **Assume a Mass for the Solution:**
- Let's assume the total mass of the solution is 100 g.
- Therefore, mass of solute A = 64 g
- Mass of solvent (water) = 100 g - 64 g = 36 g

3. **Calculate Moles of Solute A and Solvent (Water):**
- Moles of A = \( \frac{\text{mass of A}}{\text{molecular weight of A}} = \frac{64 \, \text{g}}{128 \, \text{g/mol}} = 0.5 \, \text{mol} \)
- Moles of water = \( \frac{\text{mass of water}}{\text{molecular weight of water}} = \frac{36 \, \text{g}}{18 \, \text{g/mol}} = 2 \, \text{mol} \)

4. **Calculate Mole Fractions:**
- Total moles in the solution = Moles of A + Moles of water = \( 0.5 + 2 = 2.5 \, \text{mol} \)
- Mole fraction of A, \( \chi_A = \frac{\text{moles of A}}{\text{total moles}} = \frac{0.5}{2.5} = 0.2 \)
- Mole fraction of water, \( \chi_B = 1 - \chi_A = 1 - 0.2 = 0.8 \)

5. **Apply Raoult's Law:**
- According to Raoult's Law:
\[ P_{\text{total}} = P^0_A \cdot \chi_A + P^0_{\text{H}_2\text{O}} \cdot \chi_B \]
- Rearranging to find \( P^0_A \):
\[ P^0_A = \frac{P_{\text{total}} - P^0_{\text{H}_2\text{O}} \cdot \chi_B}{\chi_A} \]

6. **Substituting the Values:**
- Substitute \( P_{\text{total}} = 145 \, \text{mmHg} \), \( P^0_{\text{H}_2\text{O}} = 155 \, \text{mmHg} \), \( \chi_B = 0.8 \), and \( \chi_A = 0.2 \):
\[ P^0_A = \frac{145 - 155 \cdot 0.8}{0.2} \]
- Calculate:
\[ P^0_A = \frac{145 - 124}{0.2} = \frac{21}{0.2} = 105 \, \text{mmHg} \]

### Final Answer:
The vapor pressure of pure liquid A is \( P^0_A = 105 \, \text{mmHg} \).