Resistance of a conductivity cell filled with ` 0.1 mol L^(-1)` KCl solution is `100Omega` . If the resistacne of the same cell when filled with 0.02 ` mol L^(-1)` KCl solution is `520 Omega` , calculate the conductivity and molar conductivity of 0.02 mol `L^(-1)` KCl solution . The conductivity of `0.1 mol L^(-1)` KCl solution is 1.29 S/m.

Explanation

The cell constant is given by the equation:
Cell constant = G = conductivity `xx` resistance
` = 1.29 S//m xx 100 Omega = 129 m^(-1) = 1.29cm^(-1)`
Conductivity of `0.02` mol `L^(-1)` KCl solution = Cell constant / resistance
` = (G)/(R) = (129 m^(-1))/(520 Omega) = 0.248 S m^(-1)`
Concentration ` = 0.02 mol L^(-1)`
` = 1000 xx 0.02 mol m^(-3)`
` = 20 mol m^(-3)`
Molare conducitivity ` = ^^_(m) = (k)/)(c)`
` = (248 xx 10^(-3) S m^(-1))/(20 mol m^(-3))`
` = 124 xx10^(-4) S m^(2) mol^(-1)`
Alternatively `k = (1.29 cm^(-1))/( 520Omega)`
` = 0.248 xx 10^(-2) S cm^(-1)`
and `^^_(m) = k xx 1000 cm^(3) L^(-1) "molarity"^(-1)`
`= 0.248 xx 10^(-2) S cm^(-1) xx 1000 cm^(3) L^(-1)/(0.02 mol L^(-1))`
` = 124 S cm^(2) mol^(-1)`