The ionisation constant of a weak electrolyte is `25xx10^(-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

`alpha=sqrt(K/C)=sqrt((25xx10^(-6))/0.01)=sqrt(25xx10^(-6))=5xx10^(-2)`
`alpha=Lambda_(eq)^c/Lambda_(eq)^oo,5xx10^(-2)=19.6/Lambda_(eq)^oo` `implies` `Lambda_(eq)^oo=19.6/(5xx10^(-2))=392` S `cm^(2) eq^(-1)`