The elctrical resistance of a column of `0.05 M N aOH ` solution of diameter `1 cm` and length `50 cm` is `5.55 xx 10^(3)ohm`. Calculate its resisteivity , conductivity, and molar conductivity.

Explanation

Cell constant , `G^(**) = (l)/(a)`
`l = 50` cm diameter = 1cm `" " therefore` radius =0.5 cm
Area of cross-section , a = `pi r^(2)`
`= 3.14 xx (0.5)^(2) = 0.785 cm^(2)`
`G^(**) = (50)/(0.785) = 63.694 cm^(-1)`
(i) Resistance , `rho = (R)/(G^(**))`
`R = 5.5 xx 10^(3) Omega`
`therefore " " rho = (5.55 xx 10^(3) Omega)/(63.694 cm^(-1)) = 87.135 Omega cm.`
(ii) Conductivity , `kappa = (1)/(rho)`
`= (1)/(87.135) = 1.148 xx 10^(-2) S cm^(-1)`
(iii) Molar conductivity , `Lambda_(m) = (kappa xx 1000)/(C)`
`C = 0.05` M
`therefore " " Lambda_(m) = (1.148 xx 10^(-2) xx 1000)/(0.05)`
` = 229.6 S cm^(2) mol^(-1)`