The boiling point elevation and freezing point depression of solutions have a number of partical applications. Ethylene glycol `(CH_(2)OH-CH_(2)OH)` is used in automobile radiatiors as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol `(CH_(3)OH)`, a fairly volatile liquid that boils only at `65^(@)C`, is sometimes used as an antifreeze in automobile radiators.
`124 g` each of the two reagents glycol and glycerol are added in `5 kg` of water of the radiators in two cars. Which of the following statements is wrong?

Options

A. Both will act as antifreeze.
B. Glycol will be better.
C. Glycerol is better because its molar mass is greater than glycol.
D. Glycol is more volatile than glycerol.

Answer

To solve the question regarding the boiling point elevation and freezing point depression of ethylene glycol and glycerol in water, we need to analyze the properties of these substances and their effects as antifreeze agents.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two substances: ethylene glycol (C₂H₆O₂) and glycerol (C₃H₈O₃), both added in equal masses (124 g) to 5 kg of water.
- We need to determine which statement about their properties as antifreeze agents is incorrect.

2. **Boiling Point Elevation and Freezing Point Depression**:
- The boiling point elevation (ΔT_b) and freezing point depression (ΔT_f) can be calculated using the formulas:
- ΔT_b = K_b * m (molality)
- ΔT_f = K_f * m (molality)
- Here, K_b and K_f are the ebullioscopic and cryoscopic constants of the solvent (water), and m is the molality of the solution.

3. **Calculating Molality**:
- Molality (m) is defined as the number of moles of solute per kilogram of solvent.
- For both ethylene glycol and glycerol:
- Molar mass of ethylene glycol (C₂H₆O₂) = 62 g/mol
- Molar mass of glycerol (C₃H₈O₃) = 92 g/mol

4. **Calculating Moles of Solute**:
- Moles of ethylene glycol = 124 g / 62 g/mol = 2 moles
- Moles of glycerol = 124 g / 92 g/mol = 1.35 moles

5. **Calculating Molality**:
- Molality of ethylene glycol = 2 moles / 5 kg = 0.4 m
- Molality of glycerol = 1.35 moles / 5 kg = 0.27 m

6. **Comparing Effects on Freezing Point and Boiling Point**:
- Since ΔT_f is directly proportional to molality, ethylene glycol will have a greater effect on lowering the freezing point compared to glycerol due to its higher molality.
- For boiling point elevation, ethylene glycol will also have a greater effect due to its higher molality.

7. **Conclusion**:
- Both ethylene glycol and glycerol are effective antifreeze agents, but ethylene glycol is more effective in both lowering the freezing point and elevating the boiling point due to its lower molar mass and higher molality.
- Therefore, any statement claiming that glycerol is better for boiling point elevation than ethylene glycol would be incorrect.