1.

If conductivity of 0.020 M KCl solution at 298 K temperature is 0.0248 S cm^(-1), then find out its molar conductivity.

Answer»

`124Omega^(-1)cm^(2)MOL^(-1)`
`224Omega^(-1)cm^(2)mol^(-1)`
`24Omega^(-1)cm^(2)mol^(-1)`
`1.24Omega^(-1)cm^(2)mol^(-1)`

Solution :Here, `K=0.0248=2.48xx10^(-2)OMEGA^(-1)cm^(-1)`
`M=0.20" mol "L^(-1)`
`Lamda_(m)=(Kxx1000)/(M)=(1000xx2.48xx10^(-2))/(0.20)`
`=(1000xx248xx10^(-2))/(20xx100)XX100`
`Lamda_(m)=124Omega^(-1)cm^(2)mol^(-1)`.


Discussion

No Comment Found

Related InterviewSolutions