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Electrochemistry

Question
CBSEENCH12006102

Electrolytic conductivity of 0.20 mol L–1 solution of KCl at 298 k is 2.48 x 10–2 ohm–1cm–1. Calculate its molar conductivity.

Solution

we have given that
electrolyitc conductivity =0.20mol/L
conductivity = 2.48 x 10-2 ohm-1cm-1
thus apply the formula

Λm= k×1000M

M=k×1000ΛmM= 2.48×10-2×10000.20

here M is molar conductivity.

Ans.  124 ohm–1 cm2 mol–1