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

Options

A. ` 50`
B. ` 196`
C. `392`
D. ` 384`

Answer

C

Explanation

`{:(HA,hArr,H^(+),+,A^(-)),(C,,0,,0),(C-Calpha,,Calpha,,Calpha):}`
Or `K= C alpha^2`
Or `25 xx 10^(-6) = 10^(-2) . alpha^2`
Or `alpha = 5 xx 10^(-2)`
also `alpha= (lambda_v)/(lambda_("infty")) = (19.6)/(lambda_("infty")) = (19.6)/(5 xx10^(-2)) = 392`.