A 0.1 molal aqueous solution of a weak acid is 30% ionised. If K f for water is 1.8 6 ∘ C / m , the freezing point of the solution will be

Freezing point depression ( Δ T ) = i K f m HA → H + + A − HA → 1 − α 1 − 0.3 i = 1 − 0.3 + 0.3 + 0.3 i = 1.3 ∴ Δ T f = 1.3 × 1.86 × 0.1 = 0.241 8 ∘ C T f = 0 − 0.241 8 ∘ C = − 0.241 8 ∘ C H + α 0.3 A − 0.3

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ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2012

ChemistryEasysingle choice

A $0.1$ molal aqueous solution of a weak acid is $30%$ ionised. If $K_{f}$ for water is $1.8 6^{\circ} C / m$ , the freezing point of the solution will be

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Solution:

Freezing point depression $(Δ T) = i K_{f} m$ $HA \to H^{+} + A -$ $HA \to 1 - α 1 - 0.3 i = 1 - 0.3 + 0.3 + 0.3 i = 1.3 ∴ Δ T_{f} = 1.3 \times 1.86 \times 0.1 = 0.241 8^{\circ} C T_{f} = 0 - 0.241 8^{\circ} C = - 0.241 8^{\circ} C H^{+} α 0.3 A^{-} 0.3$

Stream:VITEEESubject:ChemistryTopic:Liquid SolutionSubtopic:Abnormal Colligatiove Properties

⏱ 2mℹ️ Source: PYQ_2012

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