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

`(i) 258 m^(-1)`
(ii) `1.175 xx 10^(-2) S m^(2) mol^(-1)`.

Explanation

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)`