Question

Resistance of a conductivity cell (cell constant 129 m–1) filled with 74.5 ppm solution of KCI is 100 Ω (labelled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is 50 Ω (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is $\frac{\Lambda_1}{\Lambda_2} = x \times 10^{-3}$. The value of x is _________. (Nearest integer)
(Given : molar mass of KCl is 74.5 g mol–1).

Answer 1

For Solution 1,
$\Lambda_{m1} = \frac{1000K}{M}$

$M = \frac{74.5}{74.5} \times \frac{1000}{106} = 10^{-3} \, M$
[density of solution = 1 g/mol]

$\Lambda_1 = \frac{1000 \times 129 \times 10^{-4}}{10^{-3}}$

$= 129 \times 10^2 \, \text{S cm}^2 \, \text{mol}^{-1}$
$[K = \frac{x}{R} = \frac{129 \times 10^{-2}}{100}]$

For Solution 2,
$K = \frac{129 \times 10^{-2}}{50}$

$\Lambda_2 = \frac{1000 \times 129 \times 10^{-2}}{50M}$

$M = \frac{149}{74.5} \times \frac{1000}{106}$
$= 2 \times 10^{-3} \, M$

$\Lambda_2 = \frac{1000 \times 129 \times 10^{-2}}{50 \times 2 \times 10^{-3}}$
$= 129 \times 10^2 \, \text{S cm}^2 \, \text{mol}^{-1}$
$\frac{\Lambda_1}{\Lambda_2} = 1$
$\frac{\Lambda_1}{\Lambda_2} = 1000 \times 10^{-3}$
The ratio of molar conductivity of solution 1 and solution 2 is,
$\frac{\Lambda_1}{\Lambda_2} = x \times 10^{-3}$
On comparing,
$⇒ x = 1000$
So, the answer is 1000.