If glycerene `C_(3)H_(5)(OH)`, and methyl alcohol, `CH_(3)OH` are sold at the same price per kg, which would be cheaper for preparing an antifreeze solution for the radiator of an automobile ?

Answer

To determine which substance, glycerine (C₃H₅(OH)) or methyl alcohol (CH₃OH), is cheaper for preparing an antifreeze solution, we need to consider their molar masses and how they affect the freezing point depression.

### Step-by-Step Solution:

1. **Identify the Molar Masses:**
- Molar mass of glycerine (C₃H₅(OH)):
- C: 12.01 g/mol × 3 = 36.03 g/mol
- H: 1.008 g/mol × 8 = 8.064 g/mol
- O: 16.00 g/mol × 3 = 48.00 g/mol
- Total = 36.03 + 8.064 + 48.00 = 92.094 g/mol (approximately 92 g/mol)

- Molar mass of methyl alcohol (CH₃OH):
- C: 12.01 g/mol × 1 = 12.01 g/mol
- H: 1.008 g/mol × 4 = 4.032 g/mol
- O: 16.00 g/mol × 1 = 16.00 g/mol
- Total = 12.01 + 4.032 + 16.00 = 32.042 g/mol (approximately 32 g/mol)

2. **Calculate the Number of Moles in 1 kg of Each Substance:**
- For glycerine:
- Mass = 1000 g
- Moles of glycerine = Mass / Molar mass = 1000 g / 92 g/mol ≈ 10.87 moles

- For methyl alcohol:
- Mass = 1000 g
- Moles of methyl alcohol = Mass / Molar mass = 1000 g / 32 g/mol ≈ 31.25 moles

3. **Determine the Freezing Point Depression:**
- The freezing point depression (ΔTf) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(i\) = van 't Hoff factor (for both glycerine and methyl alcohol, \(i = 1\))
- \(K_f\) = freezing point depression constant (depends on the solvent, usually water)
- \(m\) = molality of the solution (moles of solute per kg of solvent)

4. **Comparison of Freezing Point Depression:**
- Since \(K_f\) is constant for water, the freezing point depression will depend on the number of moles of solute.
- Methyl alcohol has more moles (31.25 moles) compared to glycerine (10.87 moles) in 1 kg.
- Therefore, methyl alcohol will produce a greater freezing point depression than glycerine when used in equal masses.

5. **Conclusion:**
- Since both substances are sold at the same price per kg, and methyl alcohol provides a greater freezing point depression due to its higher number of moles, it will be more effective as an antifreeze solution.
- Thus, **methyl alcohol would be the cheaper option for preparing an antifreeze solution.**