The ionization constant of a weak electrolyte is `2.5 xx 10^(-5)`, while of the equivalent conductance of its `0.1 M` solution is `19.6 s cm^(2) eq^(-1)`. The equivalent conductance of the electrolyte at infinite dilution is :

Options

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

Answer

C

Explanation

`K_(a) = 25 xx 10^(-6) A_(eq) 19.6 s cm^(2) eq^(-1)`
`C = 0.1M alpha = sqrt((K_(a))/(C))=(^^_(eq)^(C))/(A_(eq)^(oo))`
`sqrt((25 xx 10^(-6))/(0.01)) = (19.6)/(^^_(eq)^(oo)) (5 xx 10^(-6))/(0.01) = (19.6)/(^^_(eq)^(oo))`
`^^_(eq)^(oo) = 392`