c) Resistance of a conductivity cell containing 0.1 M KCl solution is `100Omega`. Cell constant of the cell is 1.29/cm. Calculate the conductivity of the solution at the same temperature. 

Explanation

R of 0.1 M KCl=`100Omega("ohm")," K of 0.1 M KCl "=1.29 Sm^(-1)`
R of 0.02 M KCl `=520("ohm")`
Cell constant `="Conductivitiy " xx" Resistance "=1.29 xx 100=129 m^(-1)`
Concentration `=0.02" mol "L^(-1)=20"mol m"^(-3)`
Conductivity (k) `=("Cell constant")/("Resistance")=(129m^(-1))/(520Omega)=0.248sm^(-1)`
Molar conductivity `(wedge_(m))=K/C=(248 xx 10^(-9)Sm^(-1))/(20"mol m"^(-3))`
`=124 x 10^(-4) m^(2)"mol"^(-1)`