The temperature at which ice will begin to separate from a mixture of 20 mass percent of glycol `(C_(2)H_(6)O_(2))` in water, is : [`K_(f)` (water) = 1.86 K kg `mol^(-1)`]

Options

A. 280.5 K
B. 265. 5K
C. 276.5 K
D. 269.5 K

Answer

To solve the problem of determining the temperature at which ice will begin to separate from a mixture of 20 mass percent of glycol (C₂H₆O₂) in water, we will follow these steps:

### Step 1: Understand the mass percent
20 mass percent means that in a 100 g solution, there are 20 g of glycol (solute) and 80 g of water (solvent).

### Step 2: Calculate the number of moles of glycol
To find the number of moles of glycol, we need its molar mass. The molar mass of glycol (C₂H₆O₂) is calculated as follows:
- Carbon (C): 2 × 12 g/mol = 24 g/mol
- Hydrogen (H): 6 × 1 g/mol = 6 g/mol
- Oxygen (O): 2 × 16 g/mol = 32 g/mol

Total molar mass = 24 + 6 + 32 = 62 g/mol

Now, we can calculate the number of moles of glycol:
\[ \text{Number of moles of glycol} = \frac{\text{mass of glycol}}{\text{molar mass of glycol}} = \frac{20 \text{ g}}{62 \text{ g/mol}} \approx 0.3226 \text{ moles} \]

### Step 3: Calculate the mass of the solvent in kg
The mass of the solvent (water) is 80 g. To convert this to kg:
\[ \text{Mass of solvent} = \frac{80 \text{ g}}{1000} = 0.08 \text{ kg} \]

### Step 4: Calculate the molality (M)
Molality (M) is defined as the number of moles of solute per kg of solvent:
\[ M = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.3226 \text{ moles}}{0.08 \text{ kg}} \approx 4.0325 \text{ mol/kg} \]

### Step 5: Apply the freezing point depression formula
The freezing point depression formula is given by:
\[ \Delta T_f = i \cdot K_f \cdot M \]
Where:
- \(i\) = van 't Hoff factor (for glycol, \(i = 1\) since it does not dissociate)
- \(K_f\) = freezing point depression constant for water = 1.86 K kg/mol
- \(M\) = molality calculated above

Substituting the values:
\[ \Delta T_f = 1 \cdot 1.86 \cdot 4.0325 \approx 7.51 \text{ K} \]

### Step 6: Calculate the new freezing point
The normal freezing point of water is 273 K. The new freezing point will be:
\[ \text{New freezing point} = 273 \text{ K} - \Delta T_f = 273 \text{ K} - 7.51 \text{ K} \approx 265.49 \text{ K} \]

### Conclusion
The temperature at which ice will begin to separate from the mixture is approximately **265.5 K**.

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