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The limiting molar conductivities of HCl...

The limiting molar conductivities of HCl, `CH_(3)COONa` and NaCl are respective 425, 190 and 150 mho `cm^(2)mol^(-1)` at `25^(@)C`. The molar conductivity of 0.1 M acetic acid is 9.2 mho `cm^(2)mol^(-1)`. The degree of dissociation of 0.1 M acetic acid is

A

`0.10`

B

0.02

C

0.19

D

0.03

Text Solution

Verified by Experts

The correct Answer is:
C
`wedge_(CH_(3)COOH)^(@)=wedge_(CH_(3)COONa)^(@)+wedge_(HCl)^(@)-wedge_(NaCl)^(@)`
`" "=190+425-150=465`
`alpha=wedge_(m)^(C)/wedge_(m)^(@)=9.2/465=0.19`
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