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Question

Specific conductance of 0.1 MCH3COOH at 25 is 3.9×104ohm1cm1. If λ(H3O+) and λ(VH3COO) at 25C are 349.0 and 41.0 ohm1cm2mol1 respectively, degree of ionization of CH3COOH at the given concentration is :


A

2.0%

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B

1.0%

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C

4.0%

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D

5.0%

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Solution

The correct option is B

1.0%


λm(CH3COOH)=1000k0.1=1000×3.9×1040.1=3.9 ohm1cm2mol1αCH3COOH=λmλm=3.9λCH3COO+λH+=3.941+349=1.0×102=1%


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