I assume this is 0.1M CuSO4 and 0.1M Al2(SO4)3 IN THE SAME SOLUTION. If they are prepared solutions that have been mixed it is calculated a different way.
So it will be 0.1M in Cu^2+.
0.2 M x 2 = Al^3+
0.1M + 0.3M = 0.4M in SO4^2-
what is the total molarity of all the ions present in 0.1M cuso4 and 0.1M al2(so4)3 solution?
2 answers
Molarity=0.1+0.1+2×0.1+3×0.1
=0.1+0.1+0.2+0.3
=0.7M
=0.1+0.1+0.2+0.3
=0.7M