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Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionAbnormal Colligatiove PropertiesMedium2 minPYQ_2021

ChemistryMediumnumerical

When $9.45 g$ of ${ClCH}_{2} COOH$ is added to $500 mL$ of water, its freezing point drops by $0.5 ° C .$ The dissociation constant of ${ClCH}_{2} COOH$ is $x \times 10^{- 3} .$ The value of $x$ is off to the nearest integer)

$[K_{f (H_{2} O)} = 1.86 K kg {mol}^{- 1}]$

Answer:

36.00

Solution:

${ClCH}_{2} COOH ⇌ {ClCH}_{2} {COO}^{⊖} + H^{+}$

$i = 1 + (2 - 1) α$

$i = 1 + α$

${ΔT}_{f} = i . K_{f} . m$

$0.5 = (1 + α) (1.86) (\frac{(\frac{9.45}{94.5})}{(\frac{500}{1000})})$

$\frac{5}{3.72} = 1 + α \Rightarrow α = \frac{1.28}{3.72}$

$α = \frac{32}{93}$

$K_{a} = \frac{{(C α)}^{2}}{C - C α} = \frac{C α^{2}}{1 - α} C = \frac{0 .1}{500 / 1000} = 0 .2$

$K_{a} = \frac{0 .2 {(32 / 93)}^{2}}{(1 - 32 / 93)} = \frac{0 .2 \times {(32)}^{2}}{93 \times 61}$

$= 0 .036$

$K_{a} = 36 \times 10^{- 3}$

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

⏱ 2mℹ️ Source: PYQ_2021

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