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Question
Conductivity of 2 × 10−3 M methanoic acid is 8 × 10−5 S cm−1. Calculate its molar conductivity and degree of dissociation if `∧_"m"^0` for methanoic acid, is 404 S cm2 mol−3.
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Solution
Given: Conductivity (K) = 8 × 10−5 S cm−1
Molarity (M) = 2 × 10−3 M
Formula:
Molar conductivity, `∧_"m"^"c" = ("K" xx 1000)/"C"`
Degree of dissociation, α = `∧_"m"^"c"/∧_"m"^0`
`∧_"m"^0 = ("K" xx 1000)/"C"` S cm2 mol−1
= `(8 xx 10^-5 xx 1000)/(2 xx 10^-3)`
= 40 S cm2 mol−1
`∝` = `(∧_"m"^"c")/(∧_"m"^0)`
= `40/404`
= 0.1
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