Question

0.05 M NaOH solution offered a resistance of 31.6$\Omega$ in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm${}^{-1}$, calculate the molar conductivity of NaOH solution.

Answer 1

The relationship between conductivity $\kappa$, resistance R and the cell constant b is
$\kappa =\frac{1}{R}\times b$
$\kappa =\frac{1}{31.6\Omega }\times 0.367/cm$
$\kappa =0.0116S/cm$
The relationship between molar conductivity ${\Lambda }_m$, conductivity $\kappa$ and molar concentration C is
${\Lambda }_m=\frac{1000\kappa }{C}$
${\Lambda }_m=\frac{1000\times 0.0116S/cm}{0.05m}$
${\Lambda }_m=232.3Sc{m}^2/mol$
Hence, the molar conductivity is $232.3Sc{m}^2/mol$