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ChemistryLiquid SolutionMiscellaneous/MixedEasy2 minPYQ_2020
ChemistryEasynumerical

Depression in freezing point of 0.01 molal aqueous solution of HA is 0.01924 K. ThepKavalue of HA is
[Given = molarity = molality, K f H 2 O  = 1 .85 Kkg mol −1 , log 2 = 0.3  

Answer:
4.80
Solution:

ΔTf=i×m×Kf
0.01924 = i × 0.01 × 1.85
i = 1.04
i = 1 + α(n−1)
1.04 = 1 + α(2−1)
α=0.04
K a  =  α 2 c =  0.04 2  × 0.01 = 16 ×  10 −6
pK a  = 4.8

Stream:NTA_ABHYASSubject:ChemistryTopic:Liquid SolutionSubtopic:Miscellaneous/Mixed
⏱ 2mℹ️ Source: PYQ_2020

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