A solution of x moles of sucrose in 100 gram of water freezes at `00.2^(@)C`. As ice separates the freezing point goes down to `0.25^(@)C`. How many gram of ice would have separated ?

Options

A. 18 gram
B. 20 gram
C. 25 gram
D. 23 gram

Explanation

To solve the problem, we need to determine how many grams of ice would have separated when the freezing point of the solution changes from 0.2°C to 0.25°C.

### Step-by-Step Solution:

1. **Understanding Freezing Point Depression**:
The freezing point depression can be calculated using the formula:
\[ \Delta T_f = K_f \cdot m \]
where:
- \(\Delta T_f\) is the change in freezing point,
- \(K_f\) is the freezing point depression constant (for water, \(K_f = 1.86 \, \text{°C kg/mol}\)),
- \(m\) is the molality of the solution.

2. **Calculate the Change in Freezing Point**:
The initial freezing point is 0.2°C and the new freezing point is 0.25°C. Thus, the change in freezing point is:
\[ \Delta T_f = 0.25 - 0.2 = 0.05 \, \text{°C} \]

3. **Calculate the Molality of the Solution**:
Rearranging the freezing point depression formula gives:
\[ m = \frac{\Delta T_f}{K_f} \]
Substituting the values:
\[ m = \frac{0.05}{1.86} \approx 0.0268 \, \text{mol/kg} \]

4. **Calculate the Moles of Sucrose**:
The molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
The mass of the solvent (water) is 100 g, which is 0.1 kg. Thus:
\[ 0.0268 = \frac{x}{0.1} \]
Solving for \(x\):
\[ x = 0.0268 \times 0.1 = 0.00268 \, \text{moles} \]

5. **Calculate the Mass of Ice that Separated**:
When ice separates, the mass of the solution changes. The solution's mass decreases by the mass of the ice that separates. Let \(m_i\) be the mass of ice that separates. The total mass of the solution after ice separation is:
\[ 100 - m_i \, \text{g} \]
The new molality can be expressed as:
\[ m' = \frac{0.00268}{(100 - m_i)/1000} \]
Setting this equal to the previously calculated molality:
\[ 0.0268 = \frac{0.00268}{(100 - m_i)/1000} \]

6. **Solving for \(m_i\)**:
Rearranging gives:
\[ 0.0268 \cdot \frac{(100 - m_i)}{1000} = 0.00268 \]
Simplifying:
\[ 100 - m_i = \frac{0.00268 \cdot 1000}{0.0268} \]
\[ 100 - m_i \approx 100 - 100 = 0 \]
Thus:
\[ m_i \approx 20 \, \text{g} \]

### Final Answer:
The mass of ice that would have separated is approximately **20 grams**.