Molar Conductivity

Pattern: NEET pattern: molar conductivity problem — compute Λ_m from κ and C.

1. 1

   Given

   The specific conductance of a 0.040 M KCl solution is κ = 5.456 × 10⁻³ S cm⁻¹.

2. 2

   Required

   Calculate the molar conductivity Λ_m in S cm² mol⁻¹.

3. 3

   Concept

   Molar conductivity relates specific conductance to the amount of electrolyte. The 1000 factor accounts for volume conversion from litres to cm³.

4. 4

   Formula

   Λ_m = κ × 1000 / C

5. 5

   Substitution

   Λ_m = (5.456 × 10⁻³ S cm⁻¹) × 1000 (cm³ L⁻¹) / 0.040 (mol L⁻¹)

6. 6

   Calculation

   Numerator: 5.456 × 10⁻³ × 1000 = 5.456 S cm⁻¹ · cm³ L⁻¹ → 5.456 S cm²... wait, let me track units carefully. κ has units S cm⁻¹. Multiply by 1000 cm³ L⁻¹ gives 5.456 S cm² L⁻¹. Divide by 0.040 mol L⁻¹: Λ_m = 5.456 / 0.040 = 136.4 S cm² mol⁻¹ **Note on exact values:** The concentration 0.040 M and the factor 1000 are defined exact values (problem statement and unit conversion respectively). They do not limit significant figures. The answer precision is governed by κ (4 significant figures).

7. 7

   Final answer

   Λ_m = 136.4 S cm² mol⁻¹

8. 8

   Common trap

   The "inverts-C-multiplier" distractor: writing Λ_m = κ × C = 5.456 × 10⁻³ × 0.040 = 2.18 × 10⁻⁴ — off by a factor of ~6.25 × 10⁵ from the correct answer. Another common error is forgetting the 1000 factor, which gives 5.456 × 10⁻³ / 0.040 = 0.1364 (off by ×1000).

9. 9

   Similar NEET-style question

   "The specific conductance of 0.1 M CH₃COOH is 5.0 × 10⁻⁴ S cm⁻¹. Calculate Λ_m and, given Λ°_m = 390.5 S cm² mol⁻¹, determine the degree of dissociation α." (Answer: Λ_m = 5.0; α = Λ_m / Λ°_m = 5.0 / 390.5 ≈ 0.0128) ---