Q. The equivalent conductance of a $\text{0} \text{.2} \, \text{N}$ solution of an electrolyte was found to be $\text{200} \, \text{\Omega }^{- \text{1}} \, \text{cm}^{\text{2}} \, \text{eq}^{- \text{1}}$ . The cell constant of the cell is $\text{2} \, \text{cm}^{- 1}$ . The resistance of the solution is

NTA AbhyasNTA Abhyas 2022 Report Error

Solution:

$\text{Λ}_{\text{v}} = 200 \, \text{\Omega }^{- 1} \text{cm}^{2} \text{eq}^{- 1}$
$\text{C}_{\text{N}} = 0.2 \, \text{N,} \, \text{cm}^{- 1}$
$K=\frac{\Lambda_{v} \times C_{N}}{1000}=\frac{200 \times 0.2}{1000 \times 10}=\frac{1}{25}\Omega^{- 1}cm^{- 1}$
$R=\frac{1}{K}.\frac{l}{a}=25\times 2=50\Omega$ .