Question

A conductivity cell filled with $0.01MAgN{O}_3$ solution gives at ${25}^{o}C$ a resistance of $1442\Omega$. If the molar conductivity of $0.01MAgN{O}_3$ solution is $124.8{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$, what is the cell constant?

Answer 1

Resistance $=484\Omega$ conductivity $=\mathrm{0.00141.}{\Omega }^{-1}a{m}^{-1}$

cell constant $a=R.K$

$=484\times 0.00141=0.68244$

Now

Resistance of save conductivity cell $R=5490$

conductivity $k=\frac{G}{R}=\frac{0.68244}{5490}$

$=1.293\times {10}^{-4}$

Molar conductivity $=\frac{k}{C}$

$=\frac{1.243\times {10}^{-4}}{0.001}=0.1243$