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Electrochemistry

Question
CBSEENCH12006006

When a certain conductivity cell was filled with 0.1 M KCl, it has a resistance of 85 Q at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 Ω. Calculate the molar conductivity of the unknown electrolyte at this concentration. (Specific conductivity of 0.1 M KCl = 1.29 x 10–2 ohm–1cm–1).

Solution

Resistance of KCl solution,
         R = 85 Ω
Cell constant  = K x R
                       =1.29 ×10-2Ω-1cm-1×85Ω= 1.1 cm-1
Resistance of unknown electrolyte solution,
                  R = 96 Ω
Specific conductance
                  K = Cell constantR     = 1.1 cm-196 Ω = 11960 Ω-1 cm-1
Concentration,   
                   C = 0.052 molL    = 0.052 mol1000 cm3     = 5.2 × 10-5 mol cm-3
Molar conductance,
                Λm = KC=11 Ω-1 cm-1960×5.2×10-5 mol cm-3       = 220.2 Ω-1 cm2 mol-1.