CBSE Class 12 · 2008 · All India · Redox Reactions and ElectrochemistryShort Answer

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The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is

0.146 \times 1 0^{- 3}

S cm

^{- 1}

Answer: G* ≈ 0.219 cm⁻¹

Solution

Cell constant = conductivity × resistance:

G^{*} = κ \times R = 0.146 \times 1 0^{- 3} \times 1500

G^{*} = 0.219 c m^{- 1}

CBSE Class 12 Board · 2008 · All India · Subjective

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