Saved Bookmarks
| 1. |
Given molar conductivity of an infinite dilution: wedge_(m)^(@) for Ba(OH)_(2)=517.6Omega^(-1)cm^(2)mol^(-1). wedge_(m)^(@) for BaCl_(2)=240.6Omega^(-1)cm^(2)mol^(-1),wedge_(m)^(@) for NH_(4)Cl=129.8Omega^(-1)cm^(2)mol^(-1). Calculate wedge_(m)^(@) for NH_(4)OH. |
|
Answer» `wedge_(m)^(@)(BaCl_(2))=lamda_(Ba^(2+))^(@)+2lamda_(Cl^(-))^(@)` . . . (ii) `wedge_(m)^(@)(NH_(4)Cl)=lamda_(NH_(4)^(+))^(@)+lamda_(Cl^(-))^(@)` . . . (iii) `wedge_(m)^(@)(NH_(4)OH)=lamda_(NH_(4)^(+))^(@)+lamda_(OH^(-))^(@)+lamda_(OH^(-))^(@)=(1)/(2)eqn.(i)+eqn.(iii)-(1)/(2)` `=(1)/(2)xx517.6+129.8-(1)/(2)xx240.6=268.3Omega^(-1)cm^(2)MOL^(-1)` |
|