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The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt- electrodes. The distance between the electrodes is 120 cm with an area of cross-section of 1 c m 2 . The conductance of this solution was found to be 5 × 1 0 - 7 S. The pH of the solution is 4. The value of limiting molar conductivity Λ m o if this weak monobasic acid in aqu

Correct answer

6

Step-by-step solution

[ H + ]   =   10 − pH   =   10 − 4   M Λ M = k × 1000 M = G × l a × 1000 M = 5 × 1 0 - 7 × 120 1 × 1000 0.0015 =40   Scm 2 mol − 1 Now, H + = C α = 0.0015 × Λ m Λ m ∞ Λ m ∞ = Z × 10 2 = 0 .0015 × 40 10 − 4 On solving Z = 6.

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