Calculate the molality of aqueous solution of electrolyte that freezes at − 0.93 ∘ C if K f for water and van't Hoff factor respectively are 1.86 K kg mol − 1 and 1.25 . (Freezing point of wate

Question

Calculate the molality of aqueous solution of electrolyte that freezes at − 0.93 ∘ C if K f ​ for water and van't Hoff factor respectively are 1.86 K kg mol − 1 and 1.25 . (Freezing point of water = 0 ∘ C )

TestHub · Accepted Answer

The depression in freezing point is given by the formula:

$Δ T_{f} = i \times K_{f} \times m$

Given:

Freezing point of solution, $T_{f} = - 0.93^{\circ} C$

Freezing point of water, $T_{f}^{\circ} = 0^{\circ} C$

$Δ T_{f} = T_{f}^{\circ} - T_{f} = 0 - (- 0.93) = 0.93^{\circ} C$

van't Hoff factor, $i = 1.25$

Cryoscopic constant, $K_{f} = 1.86 K kg mol^{- 1}$

Substituting the values into the formula:

$0.93 = 1.25 \times 1.86 \times m$

$m = \frac{0.93}{1.25 \times 1.86}$

$m = \frac{0.5}{1.25} = 0.4 m$

Answer: $0.4$ m

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