A 0-05 M sodium hydroxide offered a resistance of `31.6Omega` in a conductivity cell at 298 K. Calculate the molar conductance (cell constant of cell `=0367cm^(-1)`).

Explanation

Specific conductance (k)
`=("Cell constant")/(R)`
`R=31.6Omega`
Cell constant `=0.367cm^(-1)` Specific conductance, (k)
`=(0.367)/(36.1)=0.0116Omegacm^(-1)`
Molar conductance, `Delta_(m)=(1000xxk)/(M)`
`=(1000xx0.0116)/(0.08)=232ohm^(-1)cm^(2)mol^(-1)`