CBSE Class 12 · 2004 · Delhi Compartment · Redox Reactions and ElectrochemistryShort Answer

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When a certain conductivity cell was filled with 0.1 M KCl, it had a resistance of 85 Ω at 25°C. When the same cell was filled with an aqueous solution of 0.052 M of an unknown electrolyte, the resistance was 96 Ω. Calculate the molar conductivity of the unknown electrolyte at this concentration. (Specific conductivity of 0.1 M KCl =

1.29 \times 1 0^{- 2}

ohm

^{- 1}

^{- 1}

)

Answer: Λ_m ≈ 219.7 S cm² mol⁻¹

Solution

Cell constant:

G^{*} = κ \times R = 1.29 \times 1 0^{- 2} \times 85 = 1.0965 c m^{- 1}

Conductivity of the unknown solution:

κ = \frac{G ^{*}}{R} = \frac{1.0965}{96} = 1.142 \times 1 0^{- 2} S c m^{- 1}

Molar conductivity:

Λ_{m} = \frac{κ \times 1000}{C} = \frac{1.142 \times 1 0 ^{- 2} \times 1000}{0.052}

Λ_{m} = 219.7 S c m^{2} mo l^{- 1}

CBSE Class 12 Board · 2004 · Delhi Compartment · Subjective

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CBSE Class 12 · 2004 · Delhi Compartment · Redox Reactions and ElectrochemistryShort Answer

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