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Electrochemistry

Question
CBSEENCH12006003

A 0.05 M NaOH solution offered a resistance of 31.6 Ω in a conductivity cell is 0.367 cm–1, find out the specific and molar conductance of the sodium hydroxide solution.

Solution

(i) Resistance (k) = 3.16 Ω
Conductance (C) = 1R
         = 13.16 ohm = 0.0316 ohm-1
Specific conductance (k)
                                 = Conductance x cell constant
                                = 0.0316 ohm-1×0.367 cm-1= 0.0116 ohm- cm-1
(ii) Molar conductance (C)
                         = 0.05 M = 0.05 mol L-1= 0.05 mol1 L = 0.05 mol103 cm3= 0.05 × 10-3 mol cm-3
Molar conductance (Λm)
                          = kC=(0.0116 ohm-1 cm-1)(0.05×10-3 mol cm-3)= 232 ohm-1 cm2 mol-1.