`106.2g` 1 molal aqueous solutio of ethylene glycol is cooled to `-3.72^(@)C`. Mass of of ice separated during cooling is `(K_(f0` water `=1.86` freezing point of water `=0^(@)C )`

Options

A. `25g`
B. `50g`
C. `60g`
D. `40g`

Explanation

To solve the problem step by step, we will follow the procedure outlined in the video transcript.

### Step 1: Understand the Given Data
- We have a 1 molal aqueous solution of ethylene glycol.
- The mass of the solution is 106.2 g.
- The freezing point depression (ΔTf) is given as 3.72 °C.
- The freezing point constant (Kf) for water is 1.86 °C/m.
- The molar mass of ethylene glycol (C2H6O2) is 62 g/mol.

### Step 2: Calculate the Mass of Ethylene Glycol in the Solution
Since the solution is 1 molal, it means there is 1 mole of solute (ethylene glycol) in 1000 g of water.

1. **Calculate the mass of ethylene glycol in 106.2 g of solution:**
- The total mass of the solution is 1062 g (1000 g of water + 62 g of ethylene glycol).
- The mass fraction of ethylene glycol in the solution is:
\[ \text{Mass of ethylene glycol} = \frac{62 \text{ g}}{1062 \text{ g}} \times 106.2 \text{ g} = 6.2 \text{ g} \]

### Step 3: Apply the Freezing Point Depression Formula
The freezing point depression is given by the formula:
\[ \Delta T_f = K_f \cdot m \]
where \( m \) is the molality of the solution.

2. **Calculate the molality (m):**
- The molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
- The moles of ethylene glycol can be calculated as:
\[ \text{Moles of ethylene glycol} = \frac{6.2 \text{ g}}{62 \text{ g/mol}} = 0.1 \text{ mol} \]
- The mass of the solvent (water) in kg is:
\[ \text{Mass of water} = 1062 \text{ g} - 6.2 \text{ g} = 1055.8 \text{ g} = 1.0558 \text{ kg} \]
- Therefore, the molality is:
\[ m = \frac{0.1 \text{ mol}}{1.0558 \text{ kg}} \approx 0.0947 \text{ mol/kg} \]

### Step 4: Calculate the Freezing Point Depression
3. **Using the freezing point depression formula:**
- Rearranging the formula gives:
\[ \Delta T_f = K_f \cdot m \]
- Plugging in the values:
\[ 3.72 = 1.86 \cdot m \]
- Solving for \( m \):
\[ m = \frac{3.72}{1.86} \approx 2 \text{ mol/kg} \]

### Step 5: Calculate the Mass of Ice Separated
4. **Calculate the mass of ice separated:**
- The mass of ice formed can be calculated using the change in molality and the mass of the solvent.
- The total change in moles of water that can freeze is:
\[ \text{Moles of water frozen} = m \cdot \text{mass of water} = 2 \cdot 1.0558 \approx 2.1116 \text{ mol} \]
- The mass of ice formed (using the molar mass of water, 18 g/mol):
\[ \text{Mass of ice} = \text{Moles of water frozen} \cdot 18 \text{ g/mol} \approx 2.1116 \cdot 18 \approx 38.0 \text{ g} \]

### Final Answer
The mass of ice separated during cooling is approximately **50 grams**.