The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is `5.55 xx 10^3` ohm. Calculate its resistivity, conductivity and molar conductivity.

Explanation

`R = rho l/A` or `rho =(RA)/l`
Given that: `R = 5.55 xx 10^(3) ohm, A = pir^(2) = 22/7 xx 0.5^(2) = 22/7 xx 0.25 = 0.7857 cm^(2), l = 50 cm`
Substituting the values in equation (i), we have
1. `rho = (5.55 xx 10^(3) ohm xx 0.7857 cm^(2))/(50 cm) = 87/213 ohm cm`
2. `k =1/rho =1/(87.213) ohm^(-1) cm^(-1) = 0.001147 S cm^(-1)`
3. `Lambda_(m) = k/C = (0.01147 S cm^(-1))/(0.05 mol L^(-1)) = (0.01147 S cm^(-1))/(0.05 mol (1000 cm^(3))^(-1))`
`=(0.01147 S cm^(-1))/(0.05 mol) xx 1000 cm^(3) = 229.4 S cm^(2) mol^(-1)`