2 moles of non - volatile solute is added to 1 kg water at `-8^(@)C. K_(f)` of water is 2 `"K kg mol"^(-1)`. Mass of ice that separates out is (Ignoring the effect of change in volume)

Options

A. 250 g
B. 500 g
C. 750 g
D. 100 g

Answer

To solve the problem, we need to determine the mass of ice that separates out when 2 moles of a non-volatile solute are added to 1 kg of water at -8°C. We will use the freezing point depression formula and the concept of molality.

### Step-by-Step Solution:

1. **Identify the given data:**
- Moles of solute (n) = 2 moles
- Mass of water (m_solvent) = 1 kg = 1000 g
- Freezing point depression constant (K_f) = 2 K kg mol⁻¹
- Freezing point of pure water (T_f) = 0°C
- Freezing point of solution (T_f_solution) = -8°C

2. **Calculate the depression in freezing point (ΔT_f):**
\[ \Delta T_f = T_f - T_f_solution = 0°C - (-8°C) = 8°C \]

3. **Use the freezing point depression formula:**
\[ \Delta T_f = K_f \cdot m \]
where \( m \) is the molality of the solution.

4. **Calculate the molality (m):**
Rearranging the formula gives:
\[ m = \frac{\Delta T_f}{K_f} = \frac{8°C}{2 \, \text{K kg mol}^{-1}} = 4 \, \text{mol/kg} \]

5. **Relate molality to moles of solute and mass of solvent:**
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Let \( x \) be the mass of ice that separates out. The mass of solvent (water) after ice separation is:
\[ \text{mass of solvent} = 1 \, \text{kg} - \frac{x}{1000} \, \text{kg} \]

6. **Set up the equation for molality:**
\[ 4 = \frac{2}{1 - \frac{x}{1000}} \]

7. **Cross-multiply and solve for \( x \):**
\[ 4 \left(1 - \frac{x}{1000}\right) = 2 \]
\[ 4 - \frac{4x}{1000} = 2 \]
\[ \frac{4x}{1000} = 4 - 2 = 2 \]
\[ 4x = 2000 \]
\[ x = \frac{2000}{4} = 500 \, \text{grams} \]

8. **Conclusion:**
The mass of ice that separates out is 500 grams.