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

To find the ratio of molar conductivities, we start by using the relationship between molar conductivity (\(\Lambda\)), specific conductivity (\(κ\)), and cell constant (G_cell): 
\(\Lambda = \frac{κ}{C}\), where \(C\) is the molarity.
First, convert ppm to molarity. For KCl, 74.5 ppm = 74.5 mg/L = 0.0745 g/L.
Molarity of KCl (\(M_1\)) for solution 1:
\(M_1 = \frac{0.0745 \text{ g/L}}{74.5 \text{ g/mol}} = \frac{0.0745}{74.5} \text{ mol/L} = 0.001 \text{ mol/L}\)
Similarly, for solution 2, 149 ppm = 149 mg/L = 0.149 g/L.
Molarity of KCl (\(M_2\)) for solution 2:
\(M_2 = \frac{0.149 \text{ g/L}}{74.5 \text{ g/mol}} = \frac{0.149}{74.5} \text{ mol/L} = 0.002 \text{ mol/L}\)
Next, determine specific conductivity (\(κ\)):
\(κ = \frac{G_\text{cell}}{R}\), where \(R\) is the resistance.
For solution 1,
\(κ_1 = \frac{129}{100} = 1.29 \text{ S/m}\)
For solution 2,
\(κ_2 = \frac{129}{50} = 2.58 \text{ S/m}\)
Calculate molar conductivities (\(\Lambda\)):
\(\Lambda_1 = \frac{κ_1}{M_1} = \frac{1.29}{0.001} = 1290 \text{ Scm}^2/\text{mol}\)
\(\Lambda_2 = \frac{κ_2}{M_2} = \frac{2.58}{0.002} = 1290 \text{ Scm}^2/\text{mol}\)
The ratio \(\frac{\Lambda_1}{\Lambda_2} = \frac{1290}{1290} = 1.0 = x \times 10^{-3}\)
Thus, \(x = 1000\).
Verify if \(x\) is within the range [1000, 1000]: it is.
The value of \(x\) is 1000.

Answer 2

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.