Determine the molarity and nomality of the aqueous solution containing 100g of Al2(SO4)2?

$\text{Molarity"="Moles of solute"/"Volume of solution}$
Now "moles of solute"=(100*g)/(342.15*g*mol^-1)=??*mol
And by the way $\text{aluminum sulfate}$ is $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$, the salt of $A {l}^{3 +}$ and $S {O}_{4}^{2 -}$ ions..........