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^^(m)^(0) for NaCl, HCl and NaAc are 126...

`^^_(m)^(0)` for NaCl, HCl and NaAc are `126.4, 425.0 and 91.0 S cm^(2) mol ^(-1)` respectively. Calculate `^^(0)` for Hac.

Text Solution

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`^^_(m(HAc))^(0) =lamda_(H^(+))^(0)=lamda _(Ac)^(0)=lamda_(H+)^(0) + lamda_(na^(+))^(0)-lamda_(Cl^(-))^(0) -lamda _(Na^(+))^(0)`
`=^^_(m(HCl))^(0) +^^_(m(NaAc))^(0)-^^_(m(MaCl))^(0)`
`=(425.9+ 91.0 -126.4) S cm ^(2) mol ^(-1) =390.5 S cm ^(2) mol^(-1)`
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