The molar conductivity of 0.1 M BaCl 2 solution is 106 Ω − 1 cm 2 mol − 1 at 2 5 ∘ C . What is it's conductivity?

∧ = C 1000 k ∴ k = 1000 ∧ c Now, ∧ = 106 Ω − 1 cm 2 mol − 1 , c = 0.1 mol L − 1 ∴ k = 1000 cm 3 L − 1 106 Ω − 1 cm 2 mol − 1 × 0.1 molL − 1 ∴ k = 1.06 × 1 0 − 2 Ω − 1 cm − 1

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ChemistryElectrochemistryConductance/conductivity/^mEasy2 minPYQ_2021

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The molar conductivity of $0.1 M BaCl_{2}$ solution is $106 Ω^{- 1} cm^{2}$ $mol^{- 1}$ at $2 5^{\circ} C$ . What is it's conductivity?

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Answer:

Solution:

$\land = \frac{1000 k}{C} ∴ k = \frac{\land c}{1000}$ Now, $\land = 106 Ω^{- 1} cm^{2} mol^{- 1}, c = 0.1 mol L^{- 1}$ $∴ k = \frac{106 Ω ^{- 1} cm ^{2} mol ^{- 1} \times 0.1 molL ^{- 1}}{1000 cm ^{3} L ^{- 1}} ∴ k = 1.06 \times 1 0^{- 2} Ω^{- 1} cm^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2021

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