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 conductivity \(\frac{\Lambda_1}{\Lambda_2}\) for solutions 1 and 2, we follow these steps:

1. First, calculate concentrations of KCl in mol L^–1 for both solutions. Since 1 ppm = 1 mg/L, convert ppm to g/L:
   For solution 1: 74.5 ppm = 0.0745 g/L.
   For solution 2: 149 ppm = 0.149 g/L.
2. Using the molar mass of KCl (74.5 g mol^–1), convert g/L to mol/L:
   For solution 1: \(\frac{0.0745 \text{ g/L}}{74.5 \text{ g mol}^{-1}} = 0.001 \text{ mol/L}\).
   For solution 2: \(\frac{0.149 \text{ g/L}}{74.5 \text{ g mol}^{-1}} = 0.002 \text{ mol/L}\).
3. Calculate conductivity using the formula \(\kappa = \frac{1}{R} \times \text{cell constant}\).
   For solution 1: \(\kappa_1 = \frac{1}{100 \, \Omega} \times 129 \, \text{m}^{-1} = 1.29 \, \text{S m}^{-1}\).
   For solution 2: \(\kappa_2 = \frac{1}{50 \, \Omega} \times 129 \, \text{m}^{-1} = 2.58 \, \text{S m}^{-1}\).
4. Calculate molar conductivity using \(\Lambda = \frac{\kappa}{c}\), where \(c\) is the concentration in mol/L.
   For solution 1: \(\Lambda_1 = \frac{1.29 \, \text{S m}^{-1}}{0.001 \, \text{mol/L}} = 1290 \, \text{S cm}^2 \text{ mol}^{-1}\).
   For solution 2: \(\Lambda_2 = \frac{2.58 \, \text{S m}^{-1}}{0.002 \, \text{mol/L}} = 1290 \, \text{S cm}^2 \text{ mol}^{-1}\).
5. Calculate the ratio of molar conductivities: \(\frac{\Lambda_1}{\Lambda_2} = \frac{1290}{1290} = 1\).

The ratio \(\frac{\Lambda_1}{\Lambda_2} = 1\) can be expressed as \(x \times 10^{-3}\). Solving \(x \times 10^{-3} = 1\) gives \(x = 1000\). The value \(x = 1000\) falls within the expected range (1000,1000).