0.2 molal aqueous solution of KCl freezes at − 0.68 0 ∘ C . Calculate van't Hoff factor for this solution. ( K f = 1.86 K kg mol − 1 )

Question

0.2 molal aqueous solution of KCl freezes at − 0.68 0 ∘ C . Calculate van't Hoff factor for this solution. ( K f ​ = 1.86 K kg mol − 1 )

TestHub · Accepted Answer

The depression in freezing point is given by the formula:

$Δ T_{f} = i \times K_{f} \times m$

Given:
$Δ T_{f} = 0^{\circ} C - (- 0.68 0^{\circ} C) = 0.680 K$
$K_{f} = 1.86 K kg mol^{- 1}$
$m = 0.2 m$

Substituting the values into the formula:

$0.680 = i \times 1.86 \times 0.2$

$0.680 = i \times 0.372$

$i = \frac{0.680}{0.372} \approx 1.83$

Answer: 1.83

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