Q. 22 When 1.0 g of urea (Molecular Weight = 𝟔𝟎 ) is dissolved in 200 g of solvent 𝐒, the freezing point
of 𝐒 is lowered by 𝟎. 𝟐𝟓∘𝐂. When 1.5 g of a non-electrolyte 𝐘 is dissolved in 125 g of 𝐒, the freezing point
of 𝐒 is lowered by 𝟎. 𝟐𝟎∘𝐂. The molecular weight of 𝐘 is ____ .
The correct molecular weight of solute Y is:
Final Answer
Molecular weight of Y = 180 g mol−1
Question Statement
When 1.0 g of urea (molecular weight = 60 g mol−1) is dissolved in
200 g of solvent S, the freezing point of S is lowered by
0.25 °C.
When 1.5 g of a non-electrolyte Y is dissolved in
125 g of the same solvent S, the freezing point of S is lowered by
0.20 °C.
Calculate the molecular weight of solute Y.
Concept: Freezing Point Depression
For a non-electrolyte, depression in freezing point is given by:
ΔTf = Kf × m
- ΔTf = depression in freezing point
- Kf = molal freezing point depression constant
- m = molality (mol kg−1)
Since the solvent is the same in both cases, Kf remains constant.
Step-by-Step Solution
Step 1: First Solution (Urea)
Moles of urea:
1.0 / 60 = 0.01667 mol
Mass of solvent = 200 g = 0.200 kg
Molality (m1):
m1 = 0.01667 / 0.200 = 0.08335 mol kg−1
Using ΔTf = Kf × m:
0.25 = Kf × 0.08335
Kf ≈ 3.0 °C kg mol−1
Step 2: Second Solution (Solute Y)
Let molecular weight of Y = MY
Moles of Y = 1.5 / MY
Mass of solvent = 125 g = 0.125 kg
Molality (m2):
m2 = (1.5 / MY) / 0.125 = 12 / MY
Using ΔTf = Kf × m:
0.20 = 3.0 × (12 / MY)
0.20 = 36 / MY
MY = 180 g mol−1
Explanation of MCQ Options
| Option | Status | Explanation |
|---|---|---|
| 36 g mol−1 | Incorrect | Results from incorrect solvent mass or misuse of Kf. |
| 60 g mol−1 | Incorrect | This is the molar mass of urea, often chosen by mistake. |
| 72 g mol−1 | Incorrect | Appears in similar textbook problems with different numerical data. |
| 180 g mol−1 | Correct | Matches the value obtained using correct freezing point depression calculations. |
Conclusion
Using the freezing point depression relation and given experimental data,
the correct molecular weight of solute Y is:
180 g mol−1
This problem is a classic application of
colligative properties and is highly relevant.
