2 molal solution of a weak acid HA has a freezing point of 3 . 885 ° C . The degree of dissociation of this acid is ______ × 10 - 3 . (Round off to the Nearest Integer). [ Given : Molal depression constant of water = 1 . 85 K kg mol - 1 Freezing point of pure water = 0 ° C

Δ T f = ( 1 + α ) K f × m 3 . 885 = ( 1 + α ) × 1 . 85 × 2 α = 0 . 05 = 50 × 10 - 3

Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2021

ChemistryEasynumerical

$2$ molal solution of a weak acid $HA$ has a freezing point of $3.885 ° C .$ The degree of dissociation of this acid is ______ $\times 10^{- 3} .$ (Round off to the Nearest Integer). $[$ Given : Molal depression constant of water $= 1.85 K kg {mol}^{- 1}$ Freezing point of pure water $= 0 ° C]$

Answer:

50.00

Solution:

$Δ T_{f} = (1 + α) K_{f} \times m 3.885 = (1 + α) \times 1.85 \times 2$

$α = 0.05 = 50 \times 10^{- 3}$

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

⏱ 2mℹ️ Source: PYQ_2021

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