When a solution w g of urea in `1`Kg of water is cooled to `-0.372^(@)C,200g`of ice is separated .If `K_(f)`for water is `1.86K Kgmol^(-1)`,w is

Options

A. 4.8 g
B. 12.0 g
C. 9.6 g
D. 6.0 g

Answer

To solve the problem, we need to find the weight of urea (W) in grams that is dissolved in 1 kg of water, given that when the solution is cooled to -0.372°C, 200 g of ice separates out. We will use the freezing point depression formula and the concept of molality.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Freezing point depression (ΔTf) = 0.372°C (since the freezing point of pure water is 0°C).
- Kf for water = 1.86 °C kg/mol.
- Initial weight of water (solvent) = 1 kg = 1000 g.
- Weight of ice separated = 200 g.
- Weight of solvent after ice separation = 1000 g - 200 g = 800 g = 0.800 kg.

2. **Calculate Molality (m):**
- Molality (m) is defined as the number of moles of solute per kilogram of solvent.
- We need to find the number of moles of urea (n) using the formula:
\[ \Delta Tf = Kf \times m \]
- Rearranging gives:
\[ m = \frac{\Delta Tf}{Kf} \]
- Substituting the values:
\[ m = \frac{0.372}{1.86} \approx 0.200 kg/mol \]

3. **Relate Molality to Weight of Solute:**
- Molality (m) can also be expressed as:
\[ m = \frac{n}{\text{weight of solvent in kg}} \]
- Rearranging gives:
\[ n = m \times \text{weight of solvent in kg} \]
- Substituting the values:
\[ n = 0.200 \times 0.800 = 0.160 moles \]

4. **Calculate the Weight of Urea:**
- The molecular weight of urea (NH2CONH2) is calculated as follows:
- N: 14 g/mol (2 Nitrogens = 28 g/mol)
- H: 1 g/mol (4 Hydrogens = 4 g/mol)
- C: 12 g/mol (1 Carbon = 12 g/mol)
- O: 16 g/mol (1 Oxygen = 16 g/mol)
- Total = 28 + 4 + 12 + 16 = 60 g/mol
- Now, we can find the weight of urea (W):
\[ W = n \times \text{molecular weight} \]
- Substituting the values:
\[ W = 0.160 \times 60 = 9.6 g \]

### Final Answer:
The weight of urea (W) is **9.6 g**.