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ChemistryElectrochemistryConductance/conductivity/^mMedium2 minPYQ_2014

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The equivalent conductivity of a solution containing $2.54 g$ of $CuSO_{4}$ per $L$ is $91.0 W^{- 1}$ $cm^{2} eq^{- 1}$ . Its conductivity would be

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Answer:

Solution:

We know that, $κ = Λ_{eq} \cdot C Given, Λ eq = 91.0 Ω^{- 1} cm^{2} eq^{- 1} κ = (91 Ω^{- 1} cm^{2} eq^{- 1}) = (\frac{2.54}{159/2 \times 1000} eq cm^{- 3}) = 2.9 \times 1 0^{- 3} Ω^{- 1} cm^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2014

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