0.004 M K₂SO₄ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of K₂SO₄ is ___________ (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Isotonic means equal osmotic pressure, so match iC on both sides and then read off α from i.

1. Isotonic condition. Isotonic solutions have the same osmotic pressure, π = iCRT, at the same temperature. Glucose does not dissociate, so its i = 1.

   i × 0.004 = 1 × 0.01 ⇒ i = 2.5

2. Count the ions. K₂SO₄ → 2K⁺ + SO₄²⁻ gives n = 3 ions per formula unit. For dissociation, i = 1 + (n-1)α = 1 + 2α.
3. Solve for alpha. 1 + 2α = 2.5, so α = 1.5/2 = 0.75.
4. So. The percentage dissociation of K₂SO₄ is 0.75 × 100 = 75.

The common slip: 50. Dividing by n = 3 instead of n - 1 = 2 gives α = 1.5/3 = 0.5, that is 50, not 75.

The idea to keep. For isotonic solutions, set i₁C₁ = i₂C₂. Then use i = 1 + (n-1)α, where n is the number of ions one formula unit gives.

Answer from the NTA final answer key. Solution by PrepWiser.