The vapour pressure of two miscible liquids A and B are 300 and 500 mm of Hg respectively.In a flask 2 moles of A are mixed with 6 moles of B.Further to the mixture, 32 g of an ionic non-volatile solute MCl (partically ionised mol.mass =70 u) were also added.Thus, the final vapour pressure of solution was found to be 420 mm of Hg.Then, identify the correct statement(s): (Assume the liquid mixture of A and B to behave ideally ).

Options

A. The numerical value of relative lowering in vapour pressure upon addition of solute MCl is `1//15`
B. The solute MCl is 25% ionised in the above question
C. The solute MCl is 23.33% ionised in the above question
D. Upon addition of excess `Pb(NO_3)_2`, the number of moles of `PbCl_2` precipitated is `2//35`

Explanation

To solve the problem, we will follow a systematic approach using Raoult's law and the concepts of colligative properties.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Vapor pressure of pure liquid A, \( P^0_A = 300 \, \text{mm Hg} \)
- Vapor pressure of pure liquid B, \( P^0_B = 500 \, \text{mm Hg} \)
- Moles of A, \( n_A = 2 \)
- Moles of B, \( n_B = 6 \)
- Mass of solute MCl added = 32 g
- Molar mass of MCl = 70 g/mol
- Final vapor pressure of the solution, \( P = 420 \, \text{mm Hg} \)

2. **Calculate Total Moles of the Mixture:**
\[ n_{\text{total}} = n_A + n_B = 2 + 6 = 8 \, \text{moles} \]

3. **Calculate Mole Fractions:**
- Mole fraction of A, \( X_A = \frac{n_A}{n_{\text{total}}} = \frac{2}{8} = \frac{1}{4} \)
- Mole fraction of B, \( X_B = \frac{n_B}{n_{\text{total}}} = \frac{6}{8} = \frac{3}{4} \)

4. **Calculate the Total Vapor Pressure Using Raoult's Law:**
\[ P_{\text{total}} = P^0_A \cdot X_A + P^0_B \cdot X_B \]
\[ P_{\text{total}} = 300 \cdot \frac{1}{4} + 500 \cdot \frac{3}{4} = 75 + 375 = 450 \, \text{mm Hg} \]

5. **Calculate the Relative Lowering of Vapor Pressure:**
\[ \Delta P = P_{\text{total}} - P = 450 - 420 = 30 \, \text{mm Hg} \]
\[ \text{Relative lowering of vapor pressure} = \frac{\Delta P}{P_{\text{total}}} = \frac{30}{450} = \frac{1}{15} \]

6. **Calculate Moles of Solute MCl:**
\[ \text{Moles of MCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{32 \, \text{g}}{70 \, \text{g/mol}} = \frac{32}{70} \approx 0.457 \, \text{moles} \]

7. **Calculate Total Moles in the Solution:**
\[ n_{\text{total}}' = n_{\text{total}} + \text{moles of solute} = 8 + 0.457 \approx 8.457 \, \text{moles} \]

8. **Calculate Mole Fraction of the Solute:**
\[ X_{\text{solute}} = \frac{\text{moles of solute}}{n_{\text{total}}'} = \frac{0.457}{8.457} \approx 0.054 \]

9. **Relate Relative Lowering of Vapor Pressure to Mole Fraction of Solute:**
\[ \frac{1}{15} = i \cdot X_{\text{solute}} \]
where \( i \) is the Van't Hoff factor.

10. **Calculate the Van't Hoff Factor \( i \):**
\[ i = \frac{1/15}{0.054} \approx 1.25 \]

11. **Determine Degree of Ionization:**
- For MCl, it dissociates into M\(^+\) and Cl\(^-\) (2 ions).
\[ 1 + (n - 1) \cdot \alpha = i \implies 1 + (2 - 1) \cdot \alpha = 1.25 \]
\[ \alpha = 0.25 \quad \text{(or 25% ionization)} \]

12. **Calculate Moles of Cl\(^-\) Ions:**
\[ \text{Moles of Cl}^- = \alpha \cdot \text{moles of MCl} = 0.25 \cdot 0.457 \approx 0.114 \, \text{moles} \]

13. **Calculate Moles of PbCl\(_2\) Precipitated:**
- Since each PbCl\(_2\) produces 2 Cl\(^-\):
\[ \text{Moles of PbCl}_2 = \frac{0.114}{2} \approx 0.057 \, \text{moles} \]

### Conclusion:
From the calculations, we can conclude that:
1. The numerical value of relative lowering in vapor pressure upon addition of solute MCl is \( \frac{1}{15} \) (True).
2. The solute MCl is 25% ionized (True).
3. The solute MCl is 23.33% ionized (False).
4. The number of moles of PbCl\(_2\) precipitated is \( \frac{2}{35} \) (True).

Thus, the correct statements are 1, 2, and 4.