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If the resistance of a conductivity cell containing 0.01 mol/L KCl solution at 298 K is 1000 Ω, then the conductivity of the cell is [cell constant = 0.219 cm-1] |
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Answer» Correct option is (c) As we know =1/resistance x cell constant = 0.219 cm-1 / 1000Ω = 2.19 x 10-4 5cm-1 Conductivity of 0.01 mol/ l kcl solution will be = 2.19 x 10-4 / 0.01 x 1000 = 21.9 S cm-1 |
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