The molar conductivity of 0.4 M KCl solution is 2.5 × 1 0 5 Ω − 1 cm 2 mol − 1 . What is the resistivity of solution?

Λ m = M k × 1000 2.5 × 1 0 5 = 0.4 k × 1000 Conductivity ( k ) = ρ ( resistivity ) 1 = 1000 2.5 × 1 0 5 × 0.4 = 100 Ω − 1 cm − 1 ρ = k 1 = 100 1 = 1 0 − 2 Ω cm

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ChemistryElectrochemistryConductance/conductivity/^mMedium2 minPYQ_2021

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The molar conductivity of $0.4 M KCl$ solution is $2.5 \times 1 0^{5} Ω^{- 1}$ $cm^{2} mol^{- 1}$ . What is the resistivity of solution?

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Answer:

Solution:

$Λ_{m} = \frac{k \times 1000}{M} 2.5 \times 1 0^{5} = \frac{k \times 1000}{0.4} Conductivity (k) = \frac{1}{ρ ( resistivity )} = \frac{2.5 \times 1 0 ^{5} \times 0.4}{1000} = 100 Ω^{- 1} cm^{- 1} ρ = \frac{1}{k} = \frac{1}{100} = 1 0^{- 2} Ω cm$

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

⏱ 2mℹ️ Source: PYQ_2021

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