A mixture of `(H_(2)O+C_(6)H_(5)NO_(2))` boils at `90^(@)C`. In the vapours of mixture partial vapour pressures of `H_(2)O` and `C_(6)H_(5)NO_(2)` are 733 mm Hg and 27 mm Hg respectively. The `W_(H_(2)O)//W_(C_(6)H_(5)NO_(2))` is

Explanation

To solve the problem, we need to find the weight ratio of water (H₂O) to nitrobenzene (C₆H₅NO₂) in the mixture. We can use the given partial vapor pressures and apply Raoult's Law and Dalton's Law.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Partial vapor pressure of H₂O (P₁) = 733 mmHg
- Partial vapor pressure of C₆H₅NO₂ (P₂) = 27 mmHg
- Total vapor pressure (P_total) = P₁ + P₂ = 733 mmHg + 27 mmHg = 760 mmHg

2. **Calculate Mole Fractions:**
- Mole fraction of H₂O (X₁) = P₁ / P_total = 733 mmHg / 760 mmHg
- Mole fraction of C₆H₅NO₂ (X₂) = P₂ / P_total = 27 mmHg / 760 mmHg

3. **Calculate Mole Fractions:**
- X₁ = 733 / 760 ≈ 0.9645
- X₂ = 27 / 760 ≈ 0.0355

4. **Relate Mole Fractions to Weights:**
- The mole fraction is related to the number of moles (n) and weights (W) as follows:
- X₁ = n₁ / (n₁ + n₂)
- X₂ = n₂ / (n₁ + n₂)
- The number of moles can be expressed in terms of weight and molar mass:
- n₁ = W₁ / M₁ (for H₂O, M₁ = 18 g/mol)
- n₂ = W₂ / M₂ (for C₆H₅NO₂, M₂ = 123 g/mol)

5. **Set Up the Ratios:**
- From the mole fractions:
- X₁ / X₂ = (W₁ / M₁) / (W₂ / M₂)
- Rearranging gives:
- (W₁ / W₂) = (X₁ / X₂) * (M₁ / M₂)

6. **Substituting Values:**
- Substitute the values of X₁, X₂, M₁, and M₂:
- (W₁ / W₂) = (0.9645 / 0.0355) * (18 / 123)

7. **Calculate the Weight Ratio:**
- Calculate the weight ratio:
- (W₁ / W₂) = (27.2) * (0.1463) ≈ 4.0

### Final Answer:
The weight ratio \( \frac{W_{H_2O}}{W_{C_6H_5NO_2}} \) is approximately 4.0.