Boiling point of a mixture of water and nitrobenzene is `99^(@)C`, the vapour pressure of water is `733 mm` of `Hg` and the atmospheric pressure is `760 n n` if `Hg`. The molecular weight of nitrobenzene is `123`. Find the ratio weights of the components of the distillate.

Explanation

To solve the problem, we need to find the ratio of the weights of water and nitrobenzene in the distillate. Here’s a step-by-step solution:

### Step 1: Identify the Given Data
- Boiling point of the mixture: \(99^\circ C\)
- Vapor pressure of water (\(P_{H_2O}\)): \(733 \, \text{mm Hg}\)
- Atmospheric pressure (\(P_{total}\)): \(760 \, \text{mm Hg}\)
- Molecular weight of nitrobenzene (\(M_{NB}\)): \(123 \, \text{g/mol}\)
- Molecular weight of water (\(M_{H_2O}\)): \(18 \, \text{g/mol}\)

### Step 2: Calculate the Partial Pressure of Nitrobenzene
The total pressure is the sum of the partial pressures of the components in the mixture:
\[ P_{total} = P_{H_2O} + P_{NB} \]
Rearranging gives us:
\[ P_{NB} = P_{total} - P_{H_2O} = 760 \, \text{mm Hg} - 733 \, \text{mm Hg} = 27 \, \text{mm Hg} \]

### Step 3: Set Up the Ratio of Weights
We know that the number of moles is proportional to the partial pressure. Therefore, we can express the ratio of the weights of the components as:
\[ \frac{w_{H_2O}}{w_{NB}} = \frac{P_{H_2O}}{P_{NB}} \times \frac{M_{NB}}{M_{H_2O}} \]

### Step 4: Substitute the Values
Substituting the values we have:
\[ \frac{w_{H_2O}}{w_{NB}} = \frac{733 \, \text{mm Hg}}{27 \, \text{mm Hg}} \times \frac{123 \, \text{g/mol}}{18 \, \text{g/mol}} \]

### Step 5: Calculate Each Part
1. Calculate the ratio of the partial pressures:
\[ \frac{733}{27} \approx 27.148 \]
2. Calculate the ratio of the molecular weights:
\[ \frac{123}{18} = 6.833 \]

### Step 6: Combine the Results
Now, multiply the two results:
\[ \frac{w_{H_2O}}{w_{NB}} \approx 27.148 \times 6.833 \approx 185.5 \]

### Step 7: Final Calculation
To find the ratio in a simpler form, we can express it as:
\[ \frac{w_{H_2O}}{w_{NB}} \approx 3.973 \]

### Conclusion
The ratio of the weights of water to nitrobenzene in the distillate is approximately:
\[ \frac{w_{H_2O}}{w_{NB}} \approx 3.973 \]