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ChemistryElectrochemistryConductance/conductivity/^mMedium2 minQB

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A saturated solution of AgX ( $K_{s p} = 3 \times 1 0^{- 12}$ ) and AgY ( $K_{s p} = 1 0^{- 12}$ ) has a conductivity of $0.4 \times 1 0^{- 6} Ω^{- 1} cm^{- 1}$ . Given, the limiting molar conductivity of Ag⁺ = $60 Ω^{- 1} cm^{2} mol^{- 1}$ and the limiting molar conductivity of X⁻ = $90 Ω^{- 1} cm^{2} mol^{- 1}$ .

The limiting molar conductivity of $Y^{- 1}$ is : (in $Ω^{- 1} cm^{2} mol^{- 1}$ )

Options:

Answer:

Solution:

$λ (Y^{- 1}) = \frac{K ( Y ^{-} ) \times 1000}{M ( Y ^{-} )} = \frac{1.45 \times 1 0 ^{- 7} \times 1000}{5 \times 1 0 ^{- 7}} = 290 Ω^{- 1} cm^{2} mol^{- 1}$

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

⏱ 2mℹ️ Source: QB

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