The equivalent conductivity of `KCl` at infinite dilution is `130S cm^(2) eq^(-1)`. The transport number of `Cl^(-)` ion in `KCl` at the same temperature is `0.505`. The limiting ionic mobility of `K^(+)` ion is :

Options

A. `6.67 xx 10^(-4) cm^(2) s^(-1) V^(-1)`
B. `3.22 xx 10^(-4) cm^(2) s^(-1) V^(-1)`
C. `2.20 xx 10^(-4) cm^(2) s^(-1) V^(-1)`
D. `5.01 xx 10^(-4) cm^(2) s^(-1) V^(-1)`

Answer

A

Explanation

`t_(K^(+))^(infty)=1-0.505 =0.495`
Ionic conductivity, `lamda_(K^(+))^(infty)=lamda_(KCl)^(infty)xxt_(K^(+))^(infty)`
`=130xx0.495=64.35`
`because lamda_(K^(+))^(infty)=u_(K^+)^(infty)xxF`
`:.` limiting ionic mobility,
`u_(K^(+))^(infty)=(64.35)/(96500)=6.67xx10^(-7)"cm"^(2)s^(-1)V^(-1)`.