Question

The electrodes in a conductivity cell have area $1.2\times {10}^{-4}{m}^2$ and they are fixed $3\times {10}^{-3}m$ apart. A solution containing $200g$ equivalent of the electrolyte per $m^3$ of the solution has a resistance of $60ohm$ at $298K$. Calculate the equivalent conductivity of the solution.

Answer 1

${\lambda }_{eq}=\frac{K\times 1000}{N}$

$N=200\frac{g.eq}{m^3}=200\frac{g.eq}{1000L}$

$0.2\frac{g.eq}{L}$

$k=cx=\frac{l}{RA}$

$=\frac{3\times {10}^{-3}moh{m}^{-1}}{60\times 1.2\times {10}^{-4}{m}^2}$

${\lambda }_{eq}=\frac{5}{12}\times \frac{1000}{0.2}\frac{L}{g.eq}oh{m}^{-1}{m}^{-1}$

${\lambda }_{eq}=2083.333\frac{oh{m}^{-1}{m}^{-1}}{g.eq{L}^{-1}}$