assume [Al3+]= x.then [So42-]= (3/2)x
so
x + (3/2)x = 0.1
hence
x= 0.2/5 = 0.04 M
and the apparent concentration of Al2(SO4)3 will be
(1/2)x = 0.02 M
assume [Al3+]= x.then [So42-]= (3/2)x
so
x + (3/2)x = 0.1
hence
x= 0.2/5 = 0.04 M
and the apparent concentration of Al2(SO4)3 will be
(1/2)x = 0.02 M