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Calculate Lambda(m)^(oo) for AgCl given ...

Calculate `Lambda_(m)^(oo)` for AgCl given that :
`Lambda_(m)^(oo)AgNO_(3)=133.4" ohm"^(-1)cm^(2)"equiv"^(-1)`
`Lambda_(m)^(oo)KCl=149.9" ohm"^(-1)cm^(2)"equiv"^(-1)`
`Lambda_(m)^(oo)KNO_(3)=145.1" ohm"^(-1)cm^(2)"equiv"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
`138.2" ohm^(-1) cm^(2)"equiv"^(-1)`
According to Kohirausch's law :
`Lambda_(m)^(oo)AgCl=[Lambda_(m)^(oo)AgNO_(3)]+[Lambda_(m)^(oo)KCl]-[Lambda_(m)^(oo)KNO_(3)]`
`[133.4+149.9-145.1]=138.2" ohm "^(-1)cm^(2)equiv^(-1)`.
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