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ChemistryElectrochemistryConductance/conductivity/^mMedium2 minPYQ_2022

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The conductivity of $0.20 M KCl$ solution at $300 K$ is $0.0248 Ω^{- 1} cm^{- 1}$ . What is its molar conductivity?

Answer:

Solution:

$Λ_{m} = \frac{k \times 1000}{M} = \frac{0.0248 \times 1000}{0.2} = 124 Ω^{- 1} cm^{2} mol^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2022

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