Calculate van't Hoff factor for aqueous solution of 0.02 m formic acid if it freezes at -0.045 $^{\circ} C$ .
[ $K_{f}$ = 1.86 K kg $mol^{- 1}$ and freezing point of water = $0^{\circ} C$ ]

Question

Calculate van't Hoff factor for aqueous solution of 0.02 m formic acid if it freezes at -0.045 ∘ C . [ K f ​ = 1.86 K kg mol − 1 and freezing point of water = 0 ∘ C ]

TestHub · Accepted Answer

The depression in freezing point is given by $Δ T_{f} = T_{f}^{\circ} - T_{f}$ .

$Δ T_{f} = 0 - (- 0.045) = 0.045 K$

Using the formula for depression in freezing point:

$Δ T_{f} = i \times K_{f} \times m$

Substituting the given values:

$0.045 = i \times 1.86 \times 0.02$

$0.045 = i \times 0.0372$

$i = \frac{0.045}{0.0372} = 1.209$

$i \approx 1.21$

Answer: $1.21$

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