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Question: The equivalent conductance of \(Ba^{2 +}\) and \(Cl^{-}\) are respectively 127 and \(76ohm^{- 1}cm^{...

The equivalent conductance of Ba2+Ba^{2 +} and Cl−Cl^{-} are respectively 127 and 76ohm−1cm2eq−176ohm^{- 1}cm^{2}eq^{- 1} at infinite dilution. What will be the equivalent conductance of BaCl2BaCl_{2} at infinite dilution?

A

139.ohm−1cm2eq−1139.ohm^{- 1}cm^{2}eq^{- 1}

B

203ohm−1cm2eq−1203ohm^{- 1}cm^{2}eq^{- 1}

C

279ohm−1cm2eq−1279ohm^{- 1}cm^{2}eq^{- 1}

D

101.5ohm−1cm2eq−1101.5ohm^{- 1}cm^{2}eq^{- 1}

Answer

139.ohm−1cm2eq−1139.ohm^{- 1}cm^{2}eq^{- 1}

Explanation

Solution

BaCl2→Ba2++2Cl−BaCl_{2} \rightarrow Ba^{2 +} + 2Cl^{-}

ΔºBaCl2=ΔºBa2++2ΔºCl=127+(2×76)\Delta º_{BaCl_{2}} = \Delta º_{Ba^{2 +}} + 2\Delta º_{Cl} = 127 + (2 \times 76)

=279ohm−1cm2eq−1= 279ohm^{- 1}cm^{2}eq^{- 1}

Equivalent conductivity

=2792=139.5ohm−1cm2eq−1= \frac{279}{2} = 139.5ohm^{- 1}cm^{2}eq^{- 1}