Q.17 The depression of freezing point of water (in K) for 0.1 molal solutions of NaCl and
Na2SO4 are ∆𝑇1 and ∆𝑇2, respectively. Assuming the solutions to be ideal, the
ratio ∆𝑇1 ∆𝑇2⁄ is _____________ (rounded off to two decimal places).
The ratio of freezing point depressions for 0.1 molal NaCl and Na₂SO₄ solutions is 0.67.
Freezing point depression is a colligative property depending on the number of solute particles, calculated using ΔT_f = i × K_f × m, where i is the van’t Hoff factor, K_f is water’s cryoscopic constant (1.86 K kg mol⁻¹), and m is molality. For ideal solutions at the same molality (0.1 m), the ratio ΔT₁/ΔT₂ equals i(NaCl)/i(Na₂SO₄) since K_f and m cancel out.
Van’t Hoff Factors
NaCl dissociates completely into Na⁺ + Cl⁻, yielding 2 ions, so i = 2. Na₂SO₄ dissociates into 2Na⁺ + SO₄²⁻, yielding 3 ions, so i = 3. Thus, ΔT₁ = 2 × K_f × 0.1 and ΔT₂ = 3 × K_f × 0.1.
Ratio Calculation
ΔT₁/ΔT₂ = (2 × K_f × 0.1) / (3 × K_f × 0.1) = 2/3 = 0.666…, rounded to 0.67. No options are provided in the query, but this matches standard CSIR NET expectations for ideal dissociation.
Introduction: Master Freezing Point Depression for CSIR NET
The depression of freezing point NaCl Na2SO4 0.1 molal ratio is a key colligative property question in CSIR NET Life Sciences and Chemistry exams. This problem tests understanding of van’t Hoff factors (i) in ideal solutions, where ΔT_f ∝ i × m. For 0.1 molal aqueous NaCl (ΔT₁) and Na₂SO₄ (ΔT₂), the ratio ΔT₁/ΔT₂ reveals particle count differences.
Core Concept: Colligative Properties Explained
Freezing point depression (ΔT_f) lowers pure solvent freezing by non-volatile solutes. Formula: ΔT_f = i × K_f × m (K_f water = 1.86 K kg mol⁻¹).
- i = 1 for non-electrolytes (urea).
- Electrolytes: i = total ions post-dissociation.
- Ideal solutions assume 100% dissociation, no ion pairing.
Step-by-Step Solution
Identify Dissociation:
- NaCl → Na⁺ + Cl⁻ (i = 2)
- Na₂SO₄ → 2Na⁺ + SO₄²⁻ (i = 3)
Apply Formula:
- ΔT₁ = 2 × 1.86 × 0.1 = 0.372 K
- ΔT₂ = 3 × 1.86 × 0.1 = 0.558 K
Compute Ratio:
ΔT₁/ΔT₂ = 2/3 = 0.67 (rounded to two decimals)
| Solute | i Value | ΔT_f (K) | Particles |
|---|---|---|---|
| NaCl | 2 | 0.372 | 2 ions |
| Na₂SO₄ | 3 | 0.558 | 3 ions |
CSIR NET Exam Insights
This matches GATE/CSIR patterns testing i ratios (e.g., urea:NaCl:Na₂SO₄ = 1:2:3). Real solutions show i < ideal due to ion pairing, but problem specifies “ideal”. Practice similar: 0.1 m CaCl₂ (i=3) vs. glucose (i=1).
Key Takeaways for Competitive Exams
- Ratio simplifies to i₁/i₂ for equal m.
- Na₂SO₄ causes greater depression (more particles).
- Answer: 0.67
