What is the molarity of each ion present if 18.0 g of Al2(SO4)3 is present in 201 mL of solution?

Question

_____ M Al3+

_____M SO42-

Answer 1

Molarity = moles solute / Vol Solution in Liters.

Al2(SO4)3 => 2Al^+3 + 3SO4^2-

[Al2(SO4)3] = [18/fwt Al2(SO4)3]/0.201 L

Then
[Al^+3] = 2[Molarity of Al2(SO4)3], and
[(SO4)^2-} = 3[Molarity of Al2(SO4)3]