The conductivity of 0.04 M BaCl 2 solution is 0.0112 Ω − 1 cm − 1 at 2 5 ∘ C . what is it's molar conductivity?

Λ m = M k × 1000 Conductivity ( k ) = 0.0112 Ω − 1 cm − 1 Λ m = 0.04 0.0112 × 1000 = 280 Ω − 1 cm 2 mol − 1

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ChemistryElectrochemistryConductance/conductivity/^mMedium2 minPYQ_2021

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The conductivity of $0.04 M BaCl_{2}$ solution is $0.0112 Ω^{- 1} cm^{- 1}$ at $2 5^{\circ} C$ . what is it's molar conductivity?

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Answer:

Solution:

$Λ_{m} = \frac{k \times 1000}{M}$ Conductivity $(k) = 0.0112 Ω^{- 1} cm^{- 1}$ $Λ_{m} = \frac{0.0112 \times 1000}{0.04} = 280 Ω^{- 1} cm^{2} mol^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2021

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