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Question

Specific conductance of 0.1 M HA is 3.75×104ohm1cm1. If λ(HA)=250 ohm1cm2mol1, the dissociation constant Ka of HA is :



Your Answer
A

1.0×105

Your Answer
B

2.25×104

Correct Answer
C

2.25×105

Your Answer
D

2.25×1013


Solution

The correct option is B

2.25×105


λm=1000k0.1=1000×3.75×1040.1=3.75;α=λmλm=3.75250=1.5×102;Ka=Cα2=0.1×(1.5×102)2=2.25×105

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