CBSE Class 12 · 2024 · Redox Reactions and ElectrochemistryShort Answer

The resistance of a conductivity cell filled with 0.2 mol L

^{- 1}

KCl solution is 200 Ω. If the resistance of the same cell when filled with 0.05 mol L

^{- 1}

KCl solution is 620 Ω, calculate the conductivity and molar conductivity of the 0.05 mol L

^{- 1}

KCl solution. The conductivity of 0.2 mol L

^{- 1}

KCl solution is 0.0248 S cm

^{- 1}

Answer: κ = 8.0 × 10⁻³ S cm⁻¹; Λ_m = 160 S cm² mol⁻¹

Solution

Cell constant from the 0.2 M data:

G^{*} = κ \times R = 0.0248 \times 200 = 4.96 c m^{- 1}

Conductivity of the 0.05 M solution:

κ = \frac{G ^{*}}{R} = \frac{4.96}{620} = 8.0 \times 1 0^{- 3} S c m^{- 1}

Molar conductivity:

Λ_{m} = \frac{κ \times 1000}{C} = \frac{8.0 \times 1 0 ^{- 3} \times 1000}{0.05}

Λ_{m} = 160 S c m^{2} mo l^{- 1}

CBSE Class 12 Board · 2024 · Subjective