The molar conductivities of HCl, CH₃COONa, and NaCl at infinite dilution are 426.1, 91.0 and 126.4 Ω⁻¹ cm² mol⁻¹ respectively. The molar conductivity of CH₃COOH at infinite dilution is:
A61.3 Ω⁻¹ cm² mol⁻¹
B552.5 Ω⁻¹ cm² mol⁻¹
C390.7 Ω⁻¹ cm² mol⁻¹
D200.5 Ω⁻¹ cm² mol⁻¹
Answer & Solution
Correct answer: C. 390.7 Ω⁻¹ cm² mol⁻¹
$\Lambda_0$(CH₃COOH) = $\Lambda_0$(HCl) + $\Lambda_0$(CH₃COONa) − $\Lambda_0$(NaCl) = 426.1 + 91.0 − 126.4 = **390.7** Ω⁻¹ cm² mol⁻¹. Standard Kohlrausch construction for a weak acid.
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