A 0.05 M KOH solution offered resistance of 31.6 `omega` in a conductivity cell of cel constant 0.3967 `cm^(-1)` at 298 k what is the molar conductance of KOH solution

Options

A. 150.3
B. 18068
C. 232
D. 215.7

Explanation

Given cell constant `=0.367 cm^(-1)`
Resistance =31.6 `omega` and KOH is monoacidic base so its molarity =0.05 M
k(specific conuctance ) `= "conductance" xx "cell constant"`
`=(1)/("resistance")xx("cell constant")`
`=(0.367 cm^(-1))/(31.6 omega)=0.0116 omega^(-1) cm^(-1)`
`=0.0116 S cms^(-1)`
`therefore` (molar conductance )`=(kxx1000)/(M)`
`=(kxx1000)/(0.05)=(0.0116xx1000)/(0.05)`
`=232 S cm^(2) mol^(-1)`