1.

The resistance of a conductivity cell contaning `0.001M KCl` solution at `298K` is `1500Omega`. What is the cell constant if conductivity of `0.001M KCl` solution at `298K` is `0.146 xx 10^(-3)S cm^(-3) S cm^(-1)`.

Answer» Correct Answer - `0.219cm^(-1)`
Cell constant `(G^(**))=(Conductivity(k))/(Conduct ance(G))=kxxR`
`=0.146xx10^(-3)Scm^(-1)xx1500ohm`
`=0.219cm^(-1)`


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