The ionization constant of a weak electrolyte is `25 xx 10^(-6)` while the equivalent conductance of its 0.01 M solution is 19.6 s `cm^(2) eq^(-1)`. The equivalent conductance of the electrolyte at infinite dilution (in S `cm^(2)eq^(-1)`) will be

Options

A. 250
B. 196
C. 392
D. 384

Answer

C

Explanation

`Ka=25xx10^(-6)^^_(eq)=19.6Scm^(2)eq^(-1),C=0.01`
`Ka=0.01xxalpha^(2)impliesalpha=sqrt((25xx10^(-6))/(10^(-2)))=5xx10^(-2)`
`alpha=5xx10^(-2)=(19.6)/(^^_(eq)^(@))implies^^_(eq)^(@)=(19.6)/(5xx10^(-2))=392Scm^(2)eq^(-1)`