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ChemistryElectrochemistryConductance/conductivity/^mEasy2 minPYQ_2020

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The specific conductance of a 0.5 N solution of an electrolyte at $25 ° C$ is 0.00045 $S c m^{- 1}$ . The equivalent conductance of this electrolyte at infinite dilution is 300 S $c m^{2} e q^{- 1}$ . The degree of dissociation of the electrolyte is

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Answer:

Solution:

$Λ_{v} = \frac{K \times 1000}{C_{N}} = \frac{0.00045 \times 1000}{0.5}$
$= 0.9 S c m^{2} e q^{- 1}$
$α = \frac{Λ_{v}}{Λ_{\infty}} = \frac{0.9}{300} = \frac{3}{1000} = 0.003$

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

⏱ 2mℹ️ Source: PYQ_2020

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