45 g of ethylene glycol `(C_(2) H_(6)O_(2))` is mixed with 600 g of water. The freezing point of the solution is `(K_(f)` for water is 1.86 K kg `mol^(-1)` )

Options

A. 273.95 K
B. 270.95 K
C. 370 . 95 K
D. 373.95 K

Answer

To find the freezing point of the solution when 45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water, we will use the formula for depression in freezing point:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = van't Hoff factor (for non-electrolytes like ethylene glycol, \(i = 1\))
- \(K_f\) = freezing point depression constant for water (1.86 K kg mol⁻¹)
- \(m\) = molality of the solution

### Step 1: Calculate the molar mass of ethylene glycol (C₂H₆O₂)
The molar mass can be calculated as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 6 = 6 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

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

### Step 2: Calculate the number of moles of ethylene glycol
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles of C₂H₆O₂} = \frac{45 \text{ g}}{62 \text{ g/mol}} \approx 0.7258 \text{ mol} \]

### Step 3: Calculate the mass of the solvent in kg
The mass of water is given as 600 g. To convert this to kg:
\[ \text{Mass of solvent (water)} = \frac{600 \text{ g}}{1000} = 0.6 \text{ kg} \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} \]
\[ m = \frac{0.7258 \text{ mol}}{0.6 \text{ kg}} \approx 1.2097 \text{ mol/kg} \]

### Step 5: Calculate the depression in freezing point (\(\Delta T_f\))
Now, substituting the values into the depression in freezing point formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
\[ \Delta T_f = 1 \cdot 1.86 \text{ K kg mol}^{-1} \cdot 1.2097 \text{ mol/kg} \approx 2.25 \text{ K} \]

### Step 6: Calculate the new freezing point of the solution
The normal freezing point of water is 0 °C (273.15 K). Therefore, the new freezing point will be:
\[ \text{New freezing point} = 0 - \Delta T_f = 0 - 2.25 = -2.25 \text{ °C} \]

### Final Answer
The freezing point of the solution is approximately -2.25 °C.

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