Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is / are ___________.
A. 1 M aq. NaCl and 2 M aq. urea
B. 1 M aq. CaCl₂ and 1.5 M aq. KCl
C. 1.5 M aq. AlCl₃ and 2 M aq. Na₂ SO₄
D. 2.5 M aq. KCl and 1 M aq. Al₂(SO₄)₃

Solution

Chemistry ma’am · 3 steps

Osmotic pressure counts particles, so multiply each molarity by its van't Hoff factor before comparing.

1. What isotonic means. Two solutions are isotonic when they have the same osmotic pressure. Since π = iCRT and R and T are the same for both, you only need equal values of iC.

   π₁ = π₂ ⇒ i₁ C₁ = i₂ C₂

2. Count the ions. For a strong electrolyte that dissociates fully, i is the number of ions per formula unit: NaCl and KCl give 2, CaCl₂ and Na₂SO₄ give 3, AlCl₃ gives 4, Al₂(SO₄)₃ gives 5. Urea does not dissociate, so its i = 1.

   Pair	First solution, iC	Second solution, iC	Isotonic?
   A	NaCl: 2 × 1 = 2	urea: 1 × 2 = 2	Yes
   B	CaCl₂: 3 × 1 = 3	KCl: 2 × 1.5 = 3	Yes
   C	AlCl₃: 4 × 1.5 = 6	Na₂SO₄: 3 × 2 = 6	Yes
   D	KCl: 2 × 2.5 = 5	Al₂(SO₄)₃: 5 × 1 = 5	Yes

3. So. Every pair has equal effective particle concentration, so all of them are isotonic. The number of isotonic pairs is 4.

The common slip: 0. Comparing plain molarities and ignoring the van't Hoff factor makes every pair look unequal and gives 0, not 4.

The idea to keep. For any colligative comparison, compare iC, not C. Count ions per formula unit for strong electrolytes, and take i = 1 for non-electrolytes like urea or glucose.

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