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Resistance of a conductivity cell filled with 0.1 mol L−1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L−1 - Chemistry

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Resistance of a conductivity cell filled with 0.1 mol L−1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L−1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L−1KCl solution. The conductivity of 0.1 mol L−1 KCl solution is 1.29 × 102 Ω−1 cm−1.

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Solution

Given that:
Concentration of the KCl solution = 0.1 mol L−1
Resistance of cell filled with 0.1 mol L−1 KCl solution = 100 ohm
Cell constant = G* = conductivity ×× resistance
1.29×10−2 ohm−1cm−1 × 100 ohm = 1.29 cm−1 = 129 m−1
Cell constant for a particular conductivity cell is a consant.

Conductivity of 0.02 mol L−1 KCl solution = \[\frac{Cell \ constant}{Resistance}\] \[\frac{G *}{R} = \frac{129 m^{- 1}}{520 ohm} = 0 . 248 {Sm}^{- 1}\]

Concentration = 0.02 mol L−1
= 1000××0.02 mol m−3 = 20 mol m−3
Now,

Molar conductivity = \[\Lambda_m = \frac{\kappa}{c} = \frac{248 \times {10}^{- 3} S m^{- 1}}{20 mol m^{- 3}} = 124 \times {10}^{- 4} S m^2 {mol}^{- 1}\]

Therefore, the molar conductivity of 0.02 mol L−1 KCl solution is \[124 \times {10}^{- 4} S m^2 {mol}^{- 1}\]
Concept: Conductance of Electrolytic Solutions - Introduction
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