Calculate the percent of dissociation of 0.01 m aqueous solution of K 3 [ Fe(CN) 6 ] that freezes at − 0.06 2 ∘ C. [ K f = 1.8 K kg mol − 1 ]

Question

Calculate the percent of dissociation of 0.01 m aqueous solution of K 3 ​ [ Fe(CN) 6 ​ ] that freezes at − 0.06 2 ∘ C. [ K f ​ = 1.8 K kg mol − 1 ]

TestHub · Accepted Answer

The depression in freezing point is given by the formula:

$Δ T_{f} = i \times K_{f} \times m$

Substituting the given values:

$0.062 = i \times 1.8 \times 0.01$

$i = \frac{0.062}{0.018} \approx 3.444$

The dissociation of potassium ferricyanide in aqueous solution is:

$K_{3} [Fe(CN)_{6}] \to 3 K^{+} + [Fe(CN)_{6}]^{3 -}$

The number of ions produced per formula unit, $n = 4$ .

The relationship between the van't Hoff factor ( $i$ ) and the degree of dissociation ( $α$ ) is:

$i = 1 + (n - 1) α$

$3.444 = 1 + (4 - 1) α$

$3.444 = 1 + 3 α$

$3 α = 2.444$

$α = \frac{2.444}{3} \approx 0.8147$

Therefore, the percent of dissociation is $81.47%$ .

Answer: $81.47%$

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