A solution of `x` moles of sucrose in `100`grams of water freeze at `-0.2^(@)C` As. Ice separates the freezing point goes down to `0.25^(@)C`. How many grams of ice would have separated ?

Options

A. `18`grams
B. `20`grams
C. `25`grams
D. `23`grams

Explanation

To solve the problem step by step, we will use the concept of freezing point depression.

### Step 1: Understand the given information
- We have a solution of `x` moles of sucrose in `100 g` of water.
- The freezing point of this solution is `-0.2 °C`.
- When ice separates, the freezing point goes down to `-0.25 °C`.

### Step 2: Calculate the depression in freezing point
The depression in freezing point (ΔTf) can be calculated as:
\[ \Delta T_f = T_f^{pure} - T_f^{solution} \]
Where:
- \( T_f^{pure} \) is the freezing point of pure water (0 °C).
- \( T_f^{solution} \) is the freezing point of the solution (−0.2 °C).

Thus,
\[ \Delta T_f = 0 - (-0.2) = 0.2 °C \]

### Step 3: Apply the freezing point depression formula
The formula for freezing point depression is given by:
\[ \Delta T_f = k_f \cdot m \]
Where:
- \( k_f \) is the cryoscopic constant for water (1.86 °C kg/mol).
- \( m \) is the molality of the solution.

### Step 4: Calculate the molality before ice separation
From the first scenario:
\[ 0.2 = 1.86 \cdot m \]
Solving for \( m \):
\[ m = \frac{0.2}{1.86} \approx 0.1075 \text{ mol/kg} \]

### Step 5: Calculate the number of moles of sucrose
Since molality (m) is defined as moles of solute per kg of solvent:
\[ m = \frac{x}{\text{mass of solvent in kg}} \]
Given that the mass of solvent (water) is 100 g (or 0.1 kg):
\[ 0.1075 = \frac{x}{0.1} \]
Thus,
\[ x = 0.1075 \cdot 0.1 = 0.01075 \text{ moles} \]

### Step 6: Consider the new freezing point after ice separation
After ice separates, the new freezing point is −0.25 °C. The depression in freezing point now is:
\[ \Delta T_f = 0 - (-0.25) = 0.25 °C \]

### Step 7: Set up the equation for the new molality
Using the same freezing point depression formula:
\[ 0.25 = 1.86 \cdot m' \]
Where \( m' \) is the new molality after ice has separated.

### Step 8: Calculate the new molality
Solving for \( m' \):
\[ m' = \frac{0.25}{1.86} \approx 0.1344 \text{ mol/kg} \]

### Step 9: Relate the new molality to the mass of ice separated
Let \( y \) be the mass of ice that has separated. The new mass of the solvent (water) after ice separation is \( 100 - y \) grams, or \( \frac{100 - y}{1000} \) kg.

The number of moles of sucrose remains the same (0.01075 moles), so:
\[ m' = \frac{0.01075}{\frac{100 - y}{1000}} \]
Setting this equal to the new molality:
\[ 0.1344 = \frac{0.01075}{\frac{100 - y}{1000}} \]

### Step 10: Solve for \( y \)
Rearranging gives:
\[ 0.1344 \cdot \frac{100 - y}{1000} = 0.01075 \]
\[ 100 - y = \frac{0.01075 \cdot 1000}{0.1344} \]
Calculating the right side:
\[ 100 - y \approx 79.93 \]
Thus,
\[ y \approx 100 - 79.93 \approx 20.07 \text{ grams} \]

### Conclusion
Approximately **20 grams of ice** would have separated.