A cell whose resistance, when filled with 0.1 M – KCl is 200 Ω, is measured to be 6400 Ω, when filled with 0.003 M – NaCl solution. What is the molar conductance of NaCl solution, in Ω^–1 cm^2 mol^–1

Question

A cell whose resistance, when filled with 0.1 M – KCl is 200 Ω, is measured to be 6400 Ω, when filled with 0.003 M – NaCl solution. What is the molar conductance of NaCl solution, in Ω^–1 cm^2 mol^–1 if the molar conductance of 0.1 M – KCl is 120 Ω^–1 cm^2 mol^–1?

Prashant Kotian · Accepted Answer

Correct answer: C. For 0.1 M KCl, specific conductance (k) = Molar conductance × C / 1000 = 120 × 0.1 / 1000 = 0.012 Ω^–1 cm^–1. Cell constant G* = k × R = 0.012 × 200 = 2.4 cm^–1. For 0.003 M NaCl, specific conductance (k) = G* / R = 2.4 / 6400 = 0.000375 Ω^–1 cm^–1. Molar conductance = (k × 1000) / C = (0.000375 × 1000) / 0.003 = 0.375 / 0.003 = 125 Ω^–1 cm^2 mol^–1. Therefore, correct answer is C.

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