Question:
248 g of ethylene glycol (\( C_2H_6O_2 \)) is added to 200 g of water to prepare antifreeze. What is the molality of the resultant solution?
248 g of ethylene glycol (\( C_2H_6O_2 \)) is added to 200 g of water to prepare antifreeze. What is the molality of the resultant solution?
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Always distinguish between Molality (m) and Molarity (M). Molality uses the mass of the solvent in kg, while Molarity uses the volume of the solution in liters. Molality is temperature-independent.
Updated On: Jun 3, 2026
- 5 m
- 10 m
- 20 m
- 40 m
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The Correct Option is C
Solution and Explanation
Concept:
Molality (\( m \)) is a measure of the concentration of a solute in a solution in terms of the amount of substance in a specified amount of mass of the solvent.
• Formula: \( \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \)
• Moles (\( n \)) = \( \frac{\text{Given mass}}{\text{Molar mass}} \)
Step 1: Calculating the molar mass of ethylene glycol (\( C_2H_6O_2 \)).
Atomic masses: \( C = 12, H = 1, O = 16 \) \[ \text{Molar mass} = (2 \times 12) + (6 \times 1) + (2 \times 16) \] \[ = 24 + 6 + 32 = 62 \text{ g/mol} \]
Step 2: Calculating the number of moles of solute.
Given mass of solute = 248 g \[ n = \frac{248}{62} = 4 \text{ moles} \]
Step 3: Substituting values into the molality formula.
Mass of solvent (water) = 200 g = 0.2 kg \[ m = \frac{4 \text{ moles}}{0.2 \text{ kg}} \] \[ m = \frac{40}{2} = 20 \text{ m} \]
• Formula: \( \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \)
• Moles (\( n \)) = \( \frac{\text{Given mass}}{\text{Molar mass}} \)
Step 1: Calculating the molar mass of ethylene glycol (\( C_2H_6O_2 \)).
Atomic masses: \( C = 12, H = 1, O = 16 \) \[ \text{Molar mass} = (2 \times 12) + (6 \times 1) + (2 \times 16) \] \[ = 24 + 6 + 32 = 62 \text{ g/mol} \]
Step 2: Calculating the number of moles of solute.
Given mass of solute = 248 g \[ n = \frac{248}{62} = 4 \text{ moles} \]
Step 3: Substituting values into the molality formula.
Mass of solvent (water) = 200 g = 0.2 kg \[ m = \frac{4 \text{ moles}}{0.2 \text{ kg}} \] \[ m = \frac{40}{2} = 20 \text{ m} \]
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