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The resistance of a conductivity cell filled with `0.1 M KCl` solution is `100 Omega`. If `R` of the same cell when filled with `0.02 M KCl` solution is `520 Omega`, calculate the conductivity and molar conductivity of `0.02 M KCl` solution. The conductivity of `0.1 M KCl `solution is `1.29 S m^(-1)`. |
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Answer» Step I : Let us first calculate the cell constant, Cell constant, `(G^(**))=` Conductivity `(K) xx` Resistance (R) Resistance of 0.1 M KCl solution `=100 Omega` Conductivity of 0.1 M KCl solution `=1.29 Sm^(-1)` Cell constant `=1.29 (Sm^(-1))xx100 Omega` `=129 m^(-1)` `=1.29 cm^(-1)` Step II : Calculation of conductivity of 0.02 M KCl solution Resistance of solution `=520 Omega` Cell constant `(G^(**))=1.29 cm^(-1)` Conductivity, `kappa=("Cell constant")/("Resistance")` `=(1.29 cm^(-1))/(520 Omega)` `=0.248xx10^(-2) S cm^(-1)` Step III : Calculation of molar conductivity `lambda_(m)=(1000xx kappa)/C` `C=0.02 M, kappa =0.248xx10^(-2) S cm^(-1)` `lambda_(m)=(1000xx0.248xx10^(-2))/(0.02)` `=124 S cm^(2) mol^(-1)` |
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