A weak monobasic acid is 5% dissociated in 0.01 mol `dm^(-3)` solution. The limiting molar conductivity at infinite dilution is `4.00xx10^(-2) ohm^(-1) m^(2) mol^(-1)`. Calculate the conductivity of a 0.05 mol `dm^(-3)` solution of the acid.

Answer

`8.92xx10^(-4) ohm^(-1) cm^(2) mol^(-1)`

Explanation

Dissociation constant of acid `K_(a)=Calpha^(2)`
`=0.01xx(0.05)^(2)`
`=2.5xx10^(-5)`
`K_(a)=Calpha^(2)`
`2.5xx10^(-5)=0.05xxalpha^(2)`
`alpha=0.0223`
We known that, `alpha=Lambda_(m)^(c)/Lambda_(m)^(oo)`
`0.0223=Lambda_(m)^(c)/(4xx10^(-2))`
`Lambda_(m)^(c)=8.92xx10^(-4) ohm^(-1) cm^(2) mol^(-1)`