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 finding 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 given data
- We have a solution that is 20 mass percent glycol in water.
- The freezing point depression constant (Kf) for water is given as 1.86 K kg mol⁻¹.

### Step 2: Calculate the mass of solute and solvent
- In a 100 g solution, the mass of glycol (solute) is 20 g.
- Therefore, the mass of water (solvent) is:
\[ \text{Mass of solvent} = \text{Total mass} - \text{Mass of solute} = 100 \, \text{g} - 20 \, \text{g} = 80 \, \text{g} \]

### Step 3: Convert the mass of solvent to kilograms
- Since molality is defined in kg, we convert the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent in kg} = \frac{80 \, \text{g}}{1000} = 0.08 \, \text{kg} \]

### Step 4: Calculate the molality of the solution
- The molality (m) is calculated using the formula:
\[ m = \frac{\text{mass of solute (in kg)}}{\text{molar mass of solute} \times \text{mass of solvent (in kg)}} \]
- The molar mass of glycol (C₂H₆O₂) is calculated as follows:
\[ \text{Molar mass} = (2 \times 12) + (6 \times 1) + (2 \times 16) = 24 + 6 + 32 = 62 \, \text{g/mol} \]
- Now, substituting the values into the molality formula:
\[ m = \frac{20 \, \text{g}}{62 \, \text{g/mol} \times 0.08 \, \text{kg}} = \frac{20}{62 \times 0.08} = \frac{20}{4.96} \approx 4.03 \, \text{mol/kg} \]

### Step 5: Calculate the depression in freezing point (ΔTf)
- The depression in freezing point is given by:
\[ \Delta T_f = m \times K_f \]
- Substituting the values we have:
\[ \Delta T_f = 4.03 \, \text{mol/kg} \times 1.86 \, \text{K kg/mol} \approx 7.51 \, \text{K} \]

### Step 6: Calculate the freezing point of the solution
- The freezing point of pure water is 273 K. Therefore, the freezing point of the solution is:
\[ T_f = 273 \, \text{K} - \Delta T_f = 273 \, \text{K} - 7.51 \, \text{K} \approx 265.49 \, \text{K} \]

### Step 7: Conclusion
- The temperature at which ice will begin to separate from the mixture is approximately 265.5 K.