A solution of x moles of sucrose in 100 gram of water freezes at
`0.2^(@)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 step by step, we will use the concept of freezing point depression and the formula related to it.

### Step-by-Step Solution:

**Step 1: Understand the given information.**
- We have a solution of sucrose (non-electrolyte) in 100 grams of water.
- The freezing point of the solution is initially at \(0.2^\circ C\).
- When ice separates, the freezing point drops to \(0.25^\circ C\).

**Step 2: Calculate the depression in freezing point.**
- The depression in freezing point (\(\Delta T_f\)) can be calculated as:
\[ \Delta T_f = T_f^0 - T_f \]
where \(T_f^0\) is the freezing point of pure solvent (water, \(0^\circ C\)) and \(T_f\) is the freezing point of the solution.
- Thus, for the initial freezing point:
\[ \Delta T_f = 0 - 0.2 = 0.2^\circ C \]

**Step 3: Use the freezing point depression formula.**
- The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot m \]
where \(K_f\) is the cryoscopic constant (for water, \(K_f = 1.86 \, \text{°C kg/mol}\)), and \(m\) is the molality of the solution.
- Rearranging gives us:
\[ m = \frac{\Delta T_f}{K_f} \]

**Step 4: Calculate molality using the first freezing point.**
- Substitute the values:
\[ m = \frac{0.2}{1.86} = 0.1075 \, \text{mol/kg} \]
- Since we have 100 grams of water, convert it to kilograms:
\[ \text{mass of water} = 100 \, \text{g} = 0.1 \, \text{kg} \]
- Now, using the definition of molality:
\[ m = \frac{x \, \text{moles}}{0.1 \, \text{kg}} \implies x = m \cdot 0.1 = 0.1075 \cdot 0.1 = 0.01075 \, \text{moles} \]

**Step 5: Calculate the new freezing point depression after ice separates.**
- When ice separates, the freezing point is now \(0.25^\circ C\). Calculate the new depression:
\[ \Delta T_f = 0 - 0.25 = 0.25^\circ C \]

**Step 6: Set up the equation for the new freezing point.**
- Let \(Y\) be the mass of ice that separates. The new mass of the solvent (water) will be \(100 - Y\) grams, or \((100 - Y)/1000\) kg.
- The new molality can be expressed as:
\[ m' = \frac{0.01075}{(100 - Y)/1000} \]
- Substitute into the freezing point depression formula:
\[ 0.25 = 1.86 \cdot \frac{0.01075}{(100 - Y)/1000} \]

**Step 7: Solve for \(Y\).**
- Rearranging gives:
\[ 0.25 = \frac{1.86 \cdot 0.01075 \cdot 1000}{100 - Y} \]
- Cross-multiplying and simplifying:
\[ 0.25(100 - Y) = 1.86 \cdot 0.01075 \cdot 1000 \]
\[ 25 - 0.25Y = 20.005 \]
\[ 0.25Y = 25 - 20.005 \]
\[ 0.25Y = 4.995 \implies Y = \frac{4.995}{0.25} = 19.98 \approx 20 \, \text{grams} \]

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