Question

Resistance of a $0.1MKCl$ solution in a conductance cell is $300ohm$ and specific conductance of $0.1MKCl$ is $1.33\times 10^{-2}oh{m}^{-1}c{m}^{-1}$ . The resistance of $0.1MNaCl$ solution in the same cell is $400ohm$ . The equivalent conductance of the $0.1MNaCl$ solution (in $oh{m}^{-1}c{m}^2/gmeq.$ ) is (Report the answer in three significant figures)

Answer 1

Answer: 100

$\kappa =\frac{11}{RA}$
i.e. $\frac{1}{A}=R\times \kappa =300\times 1.33\times 10^{-2}\simeq 4.0c{m}^{-1}$
${\kappa }_{NaCl}=\frac{11}{RA}=\frac{1}{400}\times 4.0$
${\Lambda }_{m(NaCl)}=\kappa \times \frac{1000}{M}=\frac{4.0}{400}\times \frac{1000}{0.1}=100oh{m}^{-1}c{m}^2/gmeq$
Here, ${\lambda }_m={\lambda }_e$