Ethylene Glycon used in "Anti freeze" has concentration of
Note - It lower the freezing point to `- 17.6^(@)`C

Options

A. 10%v/v solution
B. 20%v/v solution
C. 25%v/v solution
D. 35%v/v solution

Answer

To determine the concentration of ethylene glycol used in antifreeze that lowers the freezing point to -17.6°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Freezing Point Depression**:
- The freezing point of pure water is 0°C (273 K).
- If the freezing point is lowered to -17.6°C, we can calculate the change in temperature (ΔT).
- ΔT = Freezing point of pure water - Freezing point of solution
- ΔT = 0°C - (-17.6°C) = 17.6°C.

2. **Use the Freezing Point Depression Formula**:
- The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
- Where:
- \( \Delta T_f \) = change in freezing point (in °C)
- \( i \) = van 't Hoff factor (for ethylene glycol, \( i = 1 \) since it does not dissociate)
- \( K_f \) = freezing point depression constant for water (1.86°C kg/mol)
- \( m \) = molality of the solution.

3. **Calculate the Molality (m)**:
- Rearranging the formula gives us:
\[ m = \frac{\Delta T_f}{i \cdot K_f} \]
- Substituting the values:
\[ m = \frac{17.6°C}{1 \cdot 1.86°C \, \text{kg/mol}} \approx 9.45 \, \text{mol/kg} \]

4. **Convert Molality to Concentration**:
- Molality (m) is defined as moles of solute per kg of solvent. To find the concentration (C) in terms of volume/volume percentage, we need to know the density of the solution.
- Assuming the density of ethylene glycol is approximately 1.11 g/mL, we can convert molality to volume percentage.

5. **Determine the Volume Percentage**:
- If we have 9.45 moles of ethylene glycol in 1 kg of water (1000 g), we calculate the mass of ethylene glycol:
- Molar mass of ethylene glycol (C2H6O2) = 62.07 g/mol.
- Mass = 9.45 mol × 62.07 g/mol = 586.77 g.
- Volume of ethylene glycol = mass/density = 586.77 g / 1.11 g/mL ≈ 528.5 mL.
- Total volume of the solution = volume of ethylene glycol + volume of water (1000 mL) = 528.5 mL + 1000 mL = 1528.5 mL.
- Volume percentage = (volume of ethylene glycol / total volume) × 100 = (528.5 mL / 1528.5 mL) × 100 ≈ 34.6%.

6. **Conclusion**:
- The concentration of ethylene glycol in the antifreeze solution is approximately 35% by volume (v/v).