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Calculate lambda(m)^(@) for NH(4)OH give...

Calculate `lambda_(m)^(@)` for `NH_(4)OH` given that values of `lambda_(m)^(@)` for `Ba(OH)_(2), BaCl_(2) and NH_(4)Cl` as `523.28, 280.0` and `129.8 S cm^(2) mol^(-1)` respectively

Text Solution

Verified by Experts

`lambda_(m)^(@)NH_(4)OH=lambda_(m)^(@)+lambda_(m)^(@)(OH^(-))`
Now
`wedge_(m)^(@)NH_(4)OH=lambda_(m)^(@)NH_(4)^(@)(OH^(-))+1/2(Ba^(2+))-1/2lambda_(m)^(@)(CI^(-))-lambda_(m)^(@)(CI^(-))`
`=129.8 +1/2(523.28-280.0)=251.44 S cm^(2) mol^(-1)`
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