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:

ΔT_f = K_f × m

• ΔT_f = depression in freezing point
• K_f = molal freezing point depression constant
• m = molality (mol kg⁻¹)

Since the solvent is the same in both cases, K_f 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 ΔT_f = K_f × m:

0.25 = K_f × 0.08335

K_f ≈ 3.0 °C kg mol⁻¹

Step 2: Second Solution (Solute Y)

Let molecular weight of Y = M_Y

Moles of Y = 1.5 / M_Y

Mass of solvent = 125 g = 0.125 kg

Molality (m₂):

m₂ = (1.5 / M_Y) / 0.125 = 12 / M_Y

Using ΔT_f = K_f × m:

0.20 = 3.0 × (12 / M_Y)

0.20 = 36 / M_Y

M_Y = 180 g mol⁻¹

Explanation of MCQ Options

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.