Molar conductivity of 0.04 M BaCl 2 solution is 230 Ω − 1 cm 2 mol − 1 at 2 7 ∘ C . what is its conductivity?

Λ m = M k × 100 230 = 0.04 k × 100 k = 1000 230 × 0.04 = 9.2 × 1 0 − 3 Ω − 1 cm − 1

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ChemistryElectrochemistryConductance/conductivity/^mMedium2 minPYQ_2021

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Molar conductivity of $0.04 M BaCl_{2}$ solution is $230 Ω^{- 1} cm^{2} mol^{- 1}$ at $2 7^{\circ} C$ . what is its conductivity?

Answer:

Solution:

$Λ_{m} = \frac{k \times 100}{M} 230 = \frac{k \times 100}{0.04} k = \frac{230 \times 0.04}{1000} = 9.2 \times 1 0^{- 3} Ω^{- 1} cm^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2021

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