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Question
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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Solution
Given: C = 0.02 M
k = 2.428 × 10−3Ω−1 cm−1
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 Ω−1 cm−1 mol−1
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