Question

Specific conductivity of $N/35$ $KCl$ at $298$K is $0.002768oh{m}^{-1}c{m}^{-1}$ and it has resistance of $520$ ohm. A $N/25$ solution of a salt kept in the same cell was found to have resistance of $300$ ohm at $298$K. Calculate equivalent conductance of the solution.

Answer 1

as $R=\rho \frac{l}{A}$ $\frac{l}{A}\Rightarrow$ Cell constant

$\Rightarrow \frac{1}{R}=\frac{1}{\rho }\frac{A}{l}$ $R$= Resistance

$\Rightarrow \frac{1}{R}=K\frac{A}{l}$ $K$= Specific conductivity

$\frac{1}{520}=0.002768oh{m}^{-1}c{m}^{-1}$ , $\frac{A}{l}$

$\frac{A}{l}=0.694cm$

For salt solution of $N/25$ concentration

$R=300ohm$

$\therefore \frac{1}{R}=K^1\frac{A}{l}$

$\Rightarrow \frac{1}{300ohm}=K_1^1\left(0.694cm\right)$

$\Rightarrow K^1=4.803\times {10}^{-3}oh{m}^{-1}c{m}^{-1}$

Specific conductivity of the solution is $4.803\times {10}^{-3}oh{m}^{-1}c{m}^{-1}$

Now,

Equivalent conductivity ${\wedge }_{eq}=\frac{\text{Specific conductance (K)}\times 1000}{Normality}$

Thus, ${\wedge }_{eq}=\frac{4.803\times {10}^{-3}oh{m}^{-1}c{m}^{-1}\times 1000}{1.125}=120.075$

Thus, equivalent conductance of solution is $120.075oh{m}^{-1}c{m}^{-1}{N}^{-1}$