The value of αMΛ of Al2(SO4) is 858 Scm2 mol–1 while 024SOΛ is 160Scm2 mol–1 calculate the limiting ionic conductivity of Al3+

Dear Student,

2 λAl3+o=2 λAl3+o + λSO42-o - λSO42-o2 λAl3+o =Λm(Al2SO4)o - λSO42-o2 λAl3+o =858 - 160 =698 S cm2 mol-1λAl3+o =6982=349 S cm2 mol-1

Hence, the limiting ionic conductivity of Al3+ = 349 S cm2 mol-1

Best Wishes !

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