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The limiting equivalent conductivity of NaCl,KCl and KBr are 126.5,150.0 and 151.5 S `cm^2eq^(-1)` ,respectively. The limiting equivalent ionic conductivity for `Br^(-)` is 78 S `cm^2eq^(-1)` . The limiting equivalent ionic conductivity for `Na^+` ions would be:

A

128

B

125

C

49

D

50

Text Solution

Verified by Experts

The correct Answer is:
4
`Lambda_(m)^(oo)=(NaBr)=Lambda_(m)^(oo)(NaCl)+Lambda_(m)^(oo)(KBr)-Lambda_(m)^(oo)(KCl)`
`Lambda_(m)^(oo)(Na^(+))+Lambda_(m)^(oo)(Br)=126.5+151.5-150`
`Lambda_(m)^(oo)(Na^(+))=50`
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The limiting equivalent conductivity of NaCl, KCl and KBr are 126.5,150.0 and 151.5 S cm^(2) eq^(-1) , respectively. The limiting equivalent ionic conductance for Br^(-) is 78 S cm^(2) eq^(-1) . The limiting equivalent ionic conductance for Na^(+) ions would be :

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