Vapour pressure of pure water at `298 K ` is `23.8 mm Hg`. `50g` of urea `(NH_(2)CONH_(2))` is dissolved in `850g` of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Options

A. 23.8 mm Hg and 0.16
B. 25.4 mm Hg and 0.02
C. 30.2 mm Hg and 0.020
D. 23.4 mm Hg and 0.017

Explanation

To solve the problem of calculating the vapor pressure of water for the solution and its relative lowering, we will follow these steps:

### Step 1: Identify Given Data
- Vapor pressure of pure water, \( P_0 = 23.8 \, \text{mm Hg} \)
- Mass of urea, \( m_2 = 50 \, \text{g} \)
- Mass of water, \( m_1 = 850 \, \text{g} \)
- Molar mass of urea, \( M_2 = 60 \, \text{g/mol} \)
- Molar mass of water, \( M_1 = 18 \, \text{g/mol} \)

### Step 2: Calculate Moles of Urea and Water
1. **Moles of Urea (\( n_2 \))**:
\[ n_2 = \frac{m_2}{M_2} = \frac{50 \, \text{g}}{60 \, \text{g/mol}} = 0.8333 \, \text{mol} \]

2. **Moles of Water (\( n_1 \))**:
\[ n_1 = \frac{m_1}{M_1} = \frac{850 \, \text{g}}{18 \, \text{g/mol}} = 47.2222 \, \text{mol} \]

### Step 3: Calculate Mole Fraction of Urea
- The total number of moles in the solution is given by:
\[ n_{\text{total}} = n_1 + n_2 = 47.2222 + 0.8333 = 48.0555 \, \text{mol} \]

- Mole fraction of urea (\( X_2 \)):
\[ X_2 = \frac{n_2}{n_{\text{total}}} = \frac{0.8333}{48.0555} \approx 0.0173 \]

### Step 4: Calculate Vapor Pressure of the Solution
Using Raoult's Law, the vapor pressure of the solution (\( P \)) can be calculated as:
\[ P = P_0 \times (1 - X_2) \]

Substituting the values:
\[ P = 23.8 \, \text{mm Hg} \times (1 - 0.0173) \approx 23.8 \, \text{mm Hg} \times 0.9827 \approx 23.4 \, \text{mm Hg} \]

### Step 5: Calculate Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure is given by:
\[ \text{Relative Lowering} = \frac{P_0 - P}{P_0} = X_2 \]
From our calculations, we already found \( X_2 \approx 0.0173 \).

### Final Results
- Vapor pressure of the solution, \( P \approx 23.4 \, \text{mm Hg} \)
- Relative lowering of vapor pressure, \( \approx 0.0173 \)