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ChemistryElectrochemistryConductance/conductivity/^mEasy2 minPYQ_2020

ChemistryEasysingle choice

The limiting molar conductivities $Λ^{o}$ for NaCl, KBr and KCl are 126, 152 and $150$ $S c m^{2} m o l^{- 1}$ respectively. The $Λ^{o}$ for NaBr is

Answer:

Solution:

By kohlrausch’s law
$Λ^{o} N a B r = Λ^{o} N a C l + Λ^{o} K B r - Λ^{o} K C l$
= 126 + 152 – 150
= $128$ $S c m^{2} m o l^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2020

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