a) What mass of Al2(SO4)3 is required to make 49 mL of a 0.030-M solution of Al2(SO4)3? 0.503gb) How many moles of aluminum ions are present in the solution? for part a, i got 0.503 g Al2(SO4)3, which is correct by calculating (.03M)(342.17g)(49mL)/1000 i can't figure out how to get part b though! PLEASE HELP
It's simply a matter of converting your answer from grams to moles, starting with the 0.503g of aluminum sulfate you calculated in part ASo we have 0.503g Al2(SO4)3 and we divide that by the molar mass of the compoundThe molar mass of aluminum sulfate is 342.147 so: 0.503/342.147 0.00147 moles of Al2(SO4)3 (rounded) There are two moles of aluminum per mole of aluminum sulfate so we do: 2 x 0.00147 0.00294 moles of aluminum (rounded)
