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A conductance cell filled with 0.1 M NaCl solution has resistance 100 ohm and specific conductance 10⁻⁴ S/cm. The same cell filled with 0.01 M KCl solution has resistance 50 ohm. Calculate the molar conductance of the 0.01 M KCl solution.
- 2 S cm² / mol
- 20 S cm² / mol
- 200 S cm² / mol
- 100 S cm² / mol
Correct answer: 20 S cm² / mol
Solution
The cell constant G* = kappa(NaCl) * R(NaCl) = 10⁻⁴ * 100 = 0.01 cm⁻¹. For KCl: kappa = G*/R = 0.01/50 = 2*10⁻⁴ S/cm. With c = 0.01 mol/L = 10⁻⁵ mol/cm³, molar conductance Lambdaₘ = kappa/c = 2*10⁻⁴ / 10⁻⁵ = 20 S cm²/mol.
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