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`lamda_(m)^(0)Na^(+)=150Omega^(-1)cm^(2)"mole"^(-1)` : `lamda_(eq)^(0)Ba^(2+)=100Omega^(-1)cm^(2)aq^(-1)` `lamda_(eq)^(0)SO_(4)^(2-)=125Omega^(-1)cm^(2)eq^(-1) : `lamda_(m)^(0)Al^(3+)=300Omega^(-1)cm^(2)"mole"^(-1)` `lamda_(m)^(0)NH_(4)^(+)=200Omega^(-1)cm^(2)"mole"^(-1)` : `lamda_(m)^(0),Cl^(-)=150Omega^(-1)cm^(2)"mole"^(-1)` then calculate (a). `lamda_(eq)^(0),Al^(3+)` (b). `lamda_(eq)^(0)Al_(2)(SO_(4))^(3)` (c). `lamda_(m)^(0)(NH_(4))NaCl` (d). `lamda_(0)^(0)NaCl,BaCl_(2).6H_(2)O` (e). `lamda_(m)^(0),(NH_(4))_(2)SO_(4)Al_(2)(SO_(4))_(3).24H_(2)O` ltbr. (f). `lamda_(eq)^(0)NaCl` |
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Answer» (a) `lamda_(eq)^(0)Al^(3+)=(300)/(3)=100` (b). `lamda_(eq)^(0)Al_(2)(SO_(4))_(3)` (c). `lamda_(m)^(0)(NH_(4))_(2)SO_(4)=2xx200+2xx125=650` (d). `lamda_(m)^(0)NaCl.BaCl_(2).6H_(2)O=150+200+3xx150=800r^(-1)` (e). `lamda_(m)^(0),(NH_(4))SO_(4)Al_(2)(SO_(4))_(3).24H_(2)Olamda_(m)^(0)(NH_(4))_(2)=400=600+4xx250=2000` (f).`lamda_(eq)^(0)NaCl=300Omega^(-1)cm^(2)eq^(-1)` |
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