The number of pairs of the solutions having the same value of the osmotic pressure from the following is _________.
(Assume 100% ionization)
A. 0.500 M C₂H₅OH (aq) and 0.25 M KBr (aq)
B. 0.100 M K₄[Fe(CN)₆] (aq) and 0.100 M FeSO₄(NH₄)₂SO₄ (aq)
C. 0.05 M K₄[Fe(CN)₆] (aq) and 0.25 M NaCl (aq)
D. 0.15 M NaCl (aq) and 0.1 M BaCl₂ (aq)
E. 0.02 M KCl.MgCl₂.6H₂O (aq) and 0.05 M KCl (aq)

Solution

Chemistry ma’am · 3 steps

At the same temperature, osmotic pressure depends only on iC, so count ions and compare iC for each pair.

1. What to compare. Osmotic pressure is π = iCRT. R and T are the same for both solutions in a pair, so two solutions match only when their iC values are equal.
2. Count the ions. K₄[Fe(CN)₆] gives 4K⁺ + [Fe(CN)₆]⁴⁻, so i = 5; the complex ion stays whole. Mohr's salt gives Fe²⁺ + 2NH₄⁺ + 2SO₄²⁻, so i = 5. Carnallite, KCl.MgCl₂.6H₂O, gives K⁺ + Mg²⁺ + 3Cl⁻, so i = 5. The water of crystallisation adds no ions.

   Pair	First solution, iC	Second solution, iC	Same?
   A	C₂H₅OH: 1 × 0.5 = 0.5	KBr: 2 × 0.25 = 0.5	Yes
   B	K₄[Fe(CN)₆]: 5 × 0.1 = 0.5	Mohr's salt: 5 × 0.1 = 0.5	Yes
   C	K₄[Fe(CN)₆]: 5 × 0.05 = 0.25	NaCl: 2 × 0.25 = 0.5	No
   D	NaCl: 2 × 0.15 = 0.3	BaCl₂: 3 × 0.1 = 0.3	Yes
   E	Carnallite: 5 × 0.02 = 0.1	KCl: 2 × 0.05 = 0.1	Yes

3. So. Pairs A, B, D and E have equal iC. Only C does not match, so the answer is 4.

The common slip: 3. Breaking [Fe(CN)₆]⁴⁻ into Fe²⁺ and six CN⁻ gives i = 11, which makes pair B fail and gives 3, not 4.

The idea to keep. For colligative comparisons, multiply concentration by the number of particles. A double salt splits into all its simple ions, but a complex ion stays as one particle.

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