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The conductivity of 0.02 M AgNO3 at 25°C is 2.428 × 10−3 Ω−1 cm−1. What is its molar conductivity?

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Question

The conductivity of 0.02 M AgNO3 at 25°C is 2.428 × 103 Ω1 cm1. What is its molar conductivity?

Numerical
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Solution

Given: C = 0.02 M

k = 2.428 × 10−3Ω1 cm1

To find: Molar conductivity = ?

Formula: Molar conductivity = `(1000k)/C`

= `(1000 xx 2.428 xx 10^-3)/0.02`

= `(2425.0 xx 10^-3)/0.02`

= `2.425/0.02`

= 121.25 Ω1 cm1 mol

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2015-2016 (July)

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