The resistance of a conductivity cell with 0.1 M KCl solution is found to be 2002 at 298 K. When the same cell was filled with 0.02 M NaCl solution, the resistance at the same temperature is found to be 1100 Omega. Calculate: (i) the cell constant of the cell in m^(-1) (ii) the molar conductivity of 0.02 M NaCl solution in S m^(2)mol^(-1) Given : Conductivity of 0.1 M KCl solution at 298 K = 1.29 S m^(-1)

The resistance of a conductivity cell with 0.1 M KCl solution is found

1.	The resistance of a conductivity cell with 0.1 M KCl solution is found to be 2002 at 298 K. When the same cell was filled with 0.02 M NaCl solution, the resistance at the same temperature is found to be 1100 Omega. Calculate: (i) the cell constant of the cell in m^(-1) (ii) the molar conductivity of 0.02 M NaCl solution in S m^(2)mol^(-1) Given : Conductivity of 0.1 M KCl solution at 298 K = 1.29 S m^(-1)
Answer» Solution :Cell constant = `1.29 xx 200 = 258 m^(-1)` or `= 2.58 cm^(-1)` `kappa = 2.58 xx (1)/(1100) = 2.345 xx 10^(-2) S cm^(-1)` `LAMBDA = (2.345 xx 10^(-3) xx 1000)/(0.02) = 117.25 S cm^(2) MOL^(-1)` or `= 1.175 xx 10^(-2) S m^(2) mol^(-1)`