A KCl solution of conductivity 0.14 Sm⁻¹ shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω. The conductivity of the HCl solution is ______ *10⁻²S m⁻¹.


4 Views|Posted 5 months ago
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5 months ago

The cell constant (G) is given by G = κ * R, where κ is conductivity and R is resistance.

Since the cell constant is constant: (κ * R)KCl = (κ * R)HCl

0.14 Sm? ¹ * 4.19 Ω = κ_HCl * 1.03 Ω

κ_HCl = (0.14 * 4.19) / 1.03 = 0.569 Sm? ¹

This is equivalent to 56.9 * 10? ² Sm? ¹.

The answer, rounded off, is 57.

 

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Chemistry Hydrocarbon 2025

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