InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Calculating conductivity and molar conductivity: Resistance of a conductivity cell filled with `0.1 M KCl` solution is `100 Omega`. If the resistance of the same cell when filled with `0.02 M KCl` solutions `520 Omega` and the conductivity of `0.1 KCl` solution is `1.29 S m`, calculate the conductivity and moalr conductivity of `0.02 M KCl` solution. Strategy : Calculate the cell constant with the help of `0.01 M KCl` solution (both `R` and `kappa` are Known). Use the cell constant to determine the conductivity of `0.02 M KCl` solution and finally find its molar conductivity using the molarity |
|
Answer» Consider `0.01 M KCl` solution Cell constant `(G^(**))` = conductivity `xx` resistance `= (1.29 S m^(-1)) xx (100 Omega)` `= 1.29 Sm^(-1)` Now consider `0.02 MKCl` solution : Conductivity `(kappa)` = Cell constant `(G^(**))`/resistance `(R)` `= 129 m^(-1)//520Omega` `= 0.248 S m^(-1)` Molar concentration `= 0.02 mol L^(-1)` `= ((1000 L)/(m^(3))) xx (0.02 mol L^(-1))` `= 20 mol m^(-3)` Therefore, molar conductivity is given as `Lambda_(m) = (kappa)/(C)` `= (0.248 S m^(-1))/(20 mol m^(-3))` `= (248 xx 10^(-3) S m^(-1))/(2 xx 10^(1) mol m^(-3))` `= 124 xx 10^(-4) s m^(2)mol^(-1)` |
|