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The specific conductivity of a 0.1 M KCl solution is 1.29 × 10⁻² S/cm. The molar conductivity is

A129 S cm² mol⁻¹
B12.9 S cm² mol⁻¹
C1290 S cm² mol⁻¹
D1.29 S cm² mol⁻¹
Answer & Solution
Correct answer: A. 129 S cm² mol⁻¹
1. Molar conductivity Λ_m = κ × 1000 / c, with κ in S cm⁻¹ and c in mol L⁻¹. 2. Substitute: Λ_m = (1.29 × 10⁻²) × 1000 / 0.1. 3. = 12.9 / 0.1 = 129 S cm² mol⁻¹. 4. Units check: S cm⁻¹ × cm³ / mmol → S cm² mol⁻¹. _Source: syllabus Chemistry, Unit 3 "Electrochemistry", §3.5_
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