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Question
The specific conductance of 2.5 × 10-4 M formic acid is 5.25 × 10-5 ohm-1 cm-1. Calculate its molar conductivity and degree of dissociation.
Given `λ°_("H"^+)` = 349.5 ohm-1 cm2 mol-1 and
`λ°_("HCOO"^-) = 50.5 " ohm"^-1 "cm"^2 "mol"^-1`
Solution
Given: K = 5.25 × 10-5 ohm-1 cm-1
C = 2.5 × 10-4 M
Formula used `lambda_m = (1000 xx "K")/"C", alpha = lambda_m/lambda_0`
Molar conductivity `(lambda_m) = (1000 xx "K")/"C"`
`(lambda_m) = (100 xx 5.25 xx 10^-5)/(2.5 xx 10^-4)`
= 210 s cm2 mol-1
\[\ce{\lambda^\circ_{{\phantom{..}}CH_3COOH} = (\lambda^\circ_{\phantom{..}H} + \lambda^\circ_{\phantom{..}HCOO^-})}\]
= (349.5 + 50.5)
Degree of dissociation:
`alpha = lambda_m/lambda_0 = 210/400` = 0.525
α = 52.5%
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