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

Correct Answer: 1000

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.