A conductivity cell has its electrodes 1 cm apart and each electrode has area of cross-section 2 `cm^(2)`, when filled with M/50 solution of sodium acetate, the cell shows a resistance of 166.5 ohms. Calculate the molar conductance of sodium acetate solution at the givne concentration.

Explanation

Step I. To calculate `kappa`
`kappa=(l)/(Rxxa)`
here `l=1cm,a=2cm^(2),R=166.5Omega`
`therefore kappa=(1cm)/(166.5Omegaxx2cm^(2))=(1000xx10^(-3))/(333)S" "cm^(-1)`
`=3*00xx10^(-3)" S "cm^(-1)`
Step II. To calculate `lamda_(m)`
`lamda=(kappaxx1000cm^(3)L^(-1))/("Molarity")`
Here molarity=M/50=1/50 mol `L^(-1)`
`((3*00xx10^(-3)Scm^(-1))xx(1000cm^(3)L^(-1)))/((1//50mol" "L^(-1)))`
`=3xx50" S "cm^(2)mol^(-1)`
`=150S" "cm^(2)mol^(-1)`