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Q.

Λm for NaCl, HCl and CH3COONa are 126.4,425.9 and 91.05S cm2 mol-1 respectively. If conductivity of 0.001028 mol L-1 acetic acid solution is 4.95x 10-5S cm-1, find the degree of dissociation of the acetic acid solution

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a

1

b

0.01233

c

0.1233

d

1.233

answer is B.

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Detailed Solution

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λCH3COOH=λCH3COONa+λHClλNaCl
=425.9+91.05-126.4=390.55 
We know, Λm=k×1000M
On putting values,
λm=4.95×105×10000.001028=48.15scm2mol1α=λmλm=48.15390.55=0.1233

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