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Calculate `lambda_(m)^(@)` for `NH_(4)OH` given that values of `lambda_(m)^(@)` for `Ba(OH)_(2) BaCI_(2) and NH_(4)CI as 523.28 280.0and 129.8 S cm^(2) mol^(-1)` respectively |
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Answer» `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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