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.
If the cost of glycerol, glycol, and methanol is same, then the sequence of economy to use these compounds as antifreeze will be

Options

A. Glycerol gt Glycol gt Methanol
B. Methanol gt Glycol gt Glycerol
C. Methanol = Glycol = Glycerol
D. Methanol gt Glycol `lt` Glycerol

Explanation

To determine the sequence of economy for using glycerol, ethylene glycol, and methanol as antifreeze agents, we need to consider their effectiveness in lowering the freezing point and elevating the boiling point, which is related to their molality when dissolved in a solvent.

### Step-by-Step Solution:

1. **Understanding Freezing Point Depression and Boiling Point Elevation**:
- The freezing point depression (ΔTf) and boiling point elevation (ΔTb) are colligative properties that depend on the number of solute particles in a solution.
- The formulas are:
- ΔTf = Kf × molality
- ΔTb = Kb × molality
- Here, Kf and Kb are the freezing point depression and boiling point elevation constants, respectively.

2. **Identify the Compounds**:
- The compounds in question are:
- Ethylene glycol (C2H6O2)
- Glycerol (C3H8O3)
- Methanol (CH3OH)

3. **Molar Mass Calculation**:
- Calculate the molar mass of each compound:
- Ethylene glycol: C(2) + H(6) + O(2) = 2(12) + 6(1) + 2(16) = 62 g/mol
- Glycerol: C(3) + H(8) + O(3) = 3(12) + 8(1) + 3(16) = 92 g/mol
- Methanol: C(1) + H(4) + O(1) = 1(12) + 4(1) + 1(16) = 32 g/mol

4. **Determine Molality**:
- For a fixed mass of solute, the molality (moles of solute per kg of solvent) is inversely proportional to the molar mass of the solute.
- Therefore, the lower the molar mass, the higher the molality for the same mass of solute:
- Methanol (32 g/mol) will have the highest molality.
- Ethylene glycol (62 g/mol) will have a medium molality.
- Glycerol (92 g/mol) will have the lowest molality.

5. **Sequence of Economy**:
- Since the effectiveness of an antifreeze agent increases with molality, the sequence of economy based on their molar masses and corresponding molalities will be:
- Methanol (most economical)
- Ethylene glycol
- Glycerol (least economical)

### Final Answer:
The sequence of economy to use these compounds as antifreeze will be:
**Methanol > Ethylene Glycol > Glycerol**