Assuming complete dissociation, what is the molality of k+, br− in an aqueous solution of kbr whose freezing point is -2.53 ∘c? the molal freezing-point-depression constant of water is 1.86 (∘c⋅kg)/mol.

Dissociation of potassium bromide: KBr(aq) → K⁺(aq) + Br⁻(aq).
ΔT = 2,53°C.
Kf = 1,86°C·kg/mol.
i - Van 't Hoff factor. Because dissociate on one cation and one anions, potassium bromide has i = 2.
b - molality.
ΔT = Kf · b(solution) · i.
2,53°C = 1,86°C·kg/mol · b(solution) · 2.
b(KBr) = 0,68 m = 0,68 mol/kg.
b(KBr) = b(K⁺) = b(Br⁻) = 0,68 m.

0.68 m

Explanation:

1) The molality of the solution is found using the depression freezing-point colligative property formula:

ΔTf = i * m * Kf.

2) i is the van't Hoof factor. Which accounts for the dissociation of the KBr.

Since 1 molecule of KBr dissociates into 2 iones, 1 K+ and 1 Br-, this constant is 2., i = 2.

3) ΔTf = Tf water - Tf solution = 0 °C - (- 2.53°C) = 2.53°C

4) Kf is given: Kf = 1.86 [°C kg/mol]

5) => m = ΔTf / (i * Kf)

m = 2.53°C / (2 * 1.86°C / kg/mol) = 0.68 mol/kg = 0.68 m

6) Since, there is one mole of K+ and one mole of Br per every mole of KBr, then the molality of one of them is the same molality of the soluton

m of K+ = m of Br- = m of KBr = 0.68 m

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