0.05 M NaOH solution offered a resistance of 31.6 ohm in a conductivity cell. If the cell constant of cell is `0.367cm^(-1)`, calculate the molar conductivity of the solution.

Explanation

Molarity =0.05
`R=31.6Omega`
Cell consatant, `(l)/(a)=0.367cm^(-1)`
Molar conductivity,
`Delta_(m)=(Kxx1000)/("Molarity")`
`=((1)/(R)xx(l)/(a)xx100)/("Molarity")`
`=(1)/(31.6)xx(0.367xx1000)/(0.5)`
`=(0.367xx1000)/(31.6xx0.05)`
`=232.2ohm^(-1)cm^(2)mol^(-1)`