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?

• A. 5 m
• B. 10 m
• C. 20 m
• D. 40 m

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
**Molality (m)** is a concentration unit defined as the number of moles of solute dissolved in 1 kilogram (kg) of solvent.
It is distinct from molarity because it is based on the mass of the solvent, not the volume of the solution. This makes it independent of temperature changes.
Key Formula or Approach:
1. Molar Mass = \(\sum (\text{Atomic weights of constituent elements})\).
2. Moles of solute (\(n\)) = \(\frac{\text{Given Mass}}{\text{Molar Mass}}\).
3. Molality (\(m\)) = \(\frac{\text{Moles of Solute}}{\text{Mass of Solvent in kg}}\).
Step 2: Detailed Explanation:
1. **Find Molar Mass of Ethylene Glycol (\(C_2H_6O_2\)):**
C = 12 \(\times\) 2 = 24.
H = 1 \(\times\) 6 = 6.
O = 16 \(\times\) 2 = 32.
Molar Mass = 24 + 6 + 32 = **62 g/mol**.
2. **Calculate Moles of Solute:**
Given mass of solute = 248 g.
\(n = 248 / 62 = 4 moles\).
3. **Convert Solvent Mass to kg:**
Given mass of water (solvent) = 200 g.
In kg, mass = 200 / 1000 = **0.2 kg**.
4. **Calculate Molality:**
\(m = \text{Moles} / \text{Mass (kg)} = 4 / 0.2\).
\(m = 40 / 2 = 20 m\).
Step 3: Final Answer:
The molality of the antifreeze solution is 20 m.
This matches option (C).