A solution containing `62g` ethylene glycol in 276g water is cooled to `-10^(@)C`. If `K_(f)` for water is `1.86 kg mol^(-1)`, the amount of water (in moles) separated as ice is ________.

Explanation

To solve the problem step by step, we will follow the process of calculating the freezing point depression and then determine the amount of water that separates as ice.

### Step 1: Calculate the freezing point depression (ΔTf)
The formula for freezing point depression is given by:
\[ \Delta T_f = K_f \times m \]
Where:
- \( \Delta T_f \) = freezing point depression
- \( K_f \) = freezing point depression constant for water = 1.86 kg/mol
- \( m \) = molality of the solution

### Step 2: Determine the initial freezing point of water
The initial freezing point of pure water is 0°C. When the solution is cooled to -10°C, the freezing point depression can be calculated as:
\[ \Delta T_f = T_f^0 - T_f = 0 - (-10) = 10°C \]

### Step 3: Calculate molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the number of moles of ethylene glycol (C2H6O2).

1. **Calculate the molar mass of ethylene glycol (C2H6O2)**:
- C: 12.01 g/mol × 2 = 24.02 g/mol
- H: 1.008 g/mol × 6 = 6.048 g/mol
- O: 16.00 g/mol × 2 = 32.00 g/mol
- Total = 24.02 + 6.048 + 32.00 = 62.068 g/mol (approximately 62 g/mol)

2. **Calculate the number of moles of ethylene glycol**:
\[ \text{Moles of ethylene glycol} = \frac{62 \text{ g}}{62.068 \text{ g/mol}} \approx 1 \text{ mol} \]

3. **Convert the mass of water to kilograms**:
\[ \text{Mass of water} = 276 \text{ g} = 0.276 \text{ kg} \]

4. **Calculate molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1 \text{ mol}}{0.276 \text{ kg}} \approx 3.62 \text{ mol/kg} \]

### Step 4: Calculate the freezing point depression (ΔTf)
Now, substituting the values into the freezing point depression formula:
\[ \Delta T_f = K_f \times m = 1.86 \text{ kg/mol} \times 3.62 \text{ mol/kg} \approx 6.73°C \]

### Step 5: Determine the amount of water that remains unfrozen
Since the freezing point depression is 10°C, we need to find out how much water remains unfrozen.

1. **Calculate the amount of water that freezes**:
The total freezing point depression is 10°C, and we calculated a depression of approximately 6.73°C. The difference indicates the amount of water that remains unfrozen:
\[ \text{Amount of water that freezes} = \frac{\Delta T_f \text{ (total)}}{K_f} \times \text{mass of solvent} \]

To find the mass of water that freezes, we can use the ratio of the freezing point depression:
\[ \text{Mass of water that freezes} = \frac{10°C}{1.86°C/mol} \times 0.276 \text{ kg} \approx 1.48 \text{ kg} \]

### Step 6: Calculate the moles of water that freeze
1. **Moles of water that freeze**:
The molar mass of water (H2O) is approximately 18 g/mol.
\[ \text{Moles of water} = \frac{\text{mass of water that freezes (in grams)}}{18 \text{ g/mol}} = \frac{1480 \text{ g}}{18 \text{ g/mol}} \approx 82.22 \text{ mol} \]

### Final Answer:
The amount of water separated as ice is approximately **82.22 moles**.