मराठी

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∧m0 for methanoic acid, is 404 S cm2 mol−3. - Chemistry

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प्रश्न

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.

संख्यात्मक
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उत्तर

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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