Q.20 Solvents A and B are completely immiscible. Solute S is soluble in both these solvents. 100 g of S was added to a container which has 2kg each of A and B. The solute is 1.5 times more soluble in solvent A than in solvent B. The mixture was agitated thoroughly and allowed to reach equilibrium. Assuming that the solute has completely dissolved, the amount of solute in solvent A phase is .__________. g
Solubility in Immiscible Solvents Equilibrium Calculation
Solvents A and B, being completely immiscible, separate into distinct phases, allowing solute S to distribute between them based on relative solubility at equilibrium. Since S is 1.5 times more soluble in A than in B and fully dissolves, the distribution ratio governs how the 100 g of S partitions between the equal 2 kg volumes of each solvent. The amount of S in the A phase is 60 g.
Problem Setup
Immiscible solvents do not mix, forming two layers where solute S equilibrates independently in each based on solubility. Let solubility in B be sB g/kg; then solubility in A is sA = 1.5sB g/kg.
With 2 kg of each solvent, maximum capacity is:
- A phase: 1.5sB × 2 = 3sB g
- B phase: sB × 2 = 2sB g
- Total = 5sB g
Distribution Calculation
At equilibrium, the ratio of concentrations equals the solubility ratio:
x/2 ÷ (100−x)/2 = 1.5
Which simplifies to:
x = 1.5(100 − x)
x = 150 − 1.5x
2.5x = 150
x = 60 g in A (40 g in B)
Total capacity exceeds 100 g since 5sB > 100, where
sB = 100/5 = 20 g/kg.
Verification
The fraction going to A is:
sA /(sA + sB) = 1.5/(1.5+1) = 1.5/2.5 = 0.6
So 0.6 × 100 = 60 g, confirming equal volumes cancel out.
This assumes ideal distribution without adsorption or volume change.
Introduction to Solubility in Immiscible Solvents
In solubility in immiscible solvents, a solute partitions between non-mixing phases like water and chloroform based on relative affinities, key for extraction in biotech and GATE/IIT JAM prep. This GATE 2020 BT question exemplifies solute distribution equilibrium with 100 g S in 2 kg each of A and B, where S is 1.5x more soluble in A.
Step-by-Step Solution
Define sB as solubility in B (g/kg), so sA = 1.5sB.
Let x g dissolve in A: concentration = x/2.
In B: (100−x)/2.
Equilibrium: x/2 ÷ (100−x)/2 = 1.5 ⇒ x = 60 g.
Verify capacity: 5sB = 100 → sB = 20 g/kg, feasible.
Distribution Table
| Phase | Mass Solvent (kg) | Solubility (g/kg) | Max Capacity (g) | Equilibrium Amount (g) |
|---|---|---|---|---|
| A | 2 | 30 | 60 | 60 |
| B | 2 | 20 | 40 | 40 |
| Total | 4 | – | 100 | 100 |
Exam Relevance for IIT JAM/GATE
Ideal for biotechnology entrance exams; practice similar solute distribution problems to ace numericals. No options given—fill-in-the-blank answer is 60 g.
