The ionic conductivity of Ba2+ and Cl- at infinite dilution are 127 and 76 ohm-1 cm2 eq-1, respectively. The equivalent conductivity of BaCl2 at infinity dilution (in ohm-1cm2eq-1) would be :

BaCl2→Ba2++2Cl−ΛBaCl2∘=ΛBa2+∘+2ΛCl−∘=127+(2×76)=279 ohm−1 cm2 eq−1Equivalent conductivity =279=2792=139.5ohm−1cm2eq−1

The ionic conductivity of Ba2+ and Cl- at infinite dilution are 127 and 76 ohm-1 cm2 eq-1, respectively. The equivalent conductivity of BaCl2 at infinity dilution (in ohm-1cm2eq-1) would be :

BaCl2→Ba2++2Cl−ΛBaCl2∘=ΛBa2+∘+2ΛCl−∘=127+(2×76)=279 ohm−1 cm2 eq−1Equivalent conductivity =279=2792=139.5ohm−1cm2eq−1

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The ionic conductivity of Ba²⁺ and Cl^- at infinite dilution are 127 and 76 ohm^-1 cm²eq^-1, respectively. The equivalent conductivity of BaCl₂ at infinity dilution (in ohm^-1cm²eq^-1) would be :

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101.5

203

279

139.5

answer is D.

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Detailed Solution

$B a C l_{2} \to B a^{2 +} + 2 C l^{-}$
$Λ_{B a C l_{2}}^{\circ} = Λ_{B a^{2 +}}^{\circ} + 2 Λ_{C l^{-}}^{\circ}$
$= 127 + (2 \times 76)$
$= 279 o h m^{- 1} c m^{2} e q^{- 1}$
Equivalent conductivity =279 $= \frac{279}{2}$
=139.5ohm⁻¹cm²eq⁻¹