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ChemistryElectrochemistryConductance/conductivity/^mEasy2 minPYQ_2020

ChemistryEasysingle choice

The equivalent conductance of $1 M$ benzoic acid is $12.8 o h m^{- 1} c m^{2}$ and if the conductance of benzoate ion and $H^{+}$ ion at infinite dilution are $42$ and $288.42 o h m^{- 1} c m^{2}$ respectively. Then its degree of dissociation is

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Answer:

Solution:

$Λ_{m (C_{6} H_{5} C O O H)}^{o} = Λ_{(C_{6} H_{5} C O O^{-})}^{o} + Λ_{(H^{+})}^{o}$
$= 42 + 288.42 = 330.42$
$α = \frac{Λ_{m}^{c}}{Λ_{m}^{o}} = \frac{12.8}{330.42} = 3.9 %$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:ElectrochemistrySubtopic:Conductance/conductivity/^m

⏱ 2mℹ️ Source: PYQ_2020

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