Question

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

Answer 1

$\kappa=\frac{11}{ RA }$
i.e. $\frac{1}{ A }= R \times \kappa=300 \times 1.33 \times 10^{-2} \simeq 4.0 cm ^{-1}$
$\kappa_{ NaCl }=\frac{11}{ RA }=\frac{1}{400} \times 4.0$
$\Lambda_{ m ( NaCl )}=\kappa \times \frac{1000}{ M }=\frac{4.0}{400} \times \frac{1000}{0.1}=100 \quad ohm ^{-1} cm ^{2} / gmeq$
Here, $\lambda_{ m }=\lambda_{ e }$