`0.04 N` solution of a weak acid has specific conductance `4.23 xx 10^(-4) "mho cm"^(-1)`. If the degree of dissociation of acid at this dilution is `0.0612`, then equivalent conductivity at infinite dilution is :

Options

A. `150.8 "mho cm"^(2) eq^(-1)`
B. `172.8 "mho cm"^(2) eq^(-1)`
C. `180.6 "mho cm"^(2) eq^(-1)`
D. `160.9 "mho cm"^(2) eq^(-1)`

Answer

B

Explanation

`Lambda_(v) = kappa xx (1000)/(N)`
`= 4.23 xx 10^(-4) xx (1000)/(0.04) = 10.575`
Now `alpha = (Lambda_(v))/(Lambda_(oo))`
`:. (Lambda_(oo)) = (10.575)/(0.0612) = 172.79`