a) How many moles of Al 2 (SO 4 ) 3 are required to make 48 mL of a 0.010 M Al 2
ID: 934621 • Letter: A
Question
a) How many moles of Al2(SO4)3 are required to make 48 mL of a 0.010 M Al2(SO4)3 solution?
(Hint, molarity is moles per liter. You know the molarity. You are given the number of mL's, which can be converted to liters. Just set it up so the units cancel and you get "moles".)
moles Al2(SO4)3
b) What is the molecular weight of Al2(SO4)3, to the nearest gram?
grams
c) What mass of Al2(SO4)3 is required to make 48 mL of a 0.010 M Al2(SO4)3 solution?
(If you've got moles and molecular weight you can get to grams.)
g Al2(SO4)3
d) How many moles of aluminum ions are present in the solution?
(This is easy. You know how many moles of Al2(SO4)3 there are from part a. Just think about how many aluminum ions there are per Al2(SO4)3.)
mol Al3+
Explanation / Answer
a.Al2(SO4)3
you want 0.010M but only 0.048 liter
so you need 0.010 moles/liter = x moles/0.048liter
xmoles = 0.00048 moles
b. (MW= 342.15 g/mol)
c. 0.00048moles * 342.15 g/mole = 0.164232 g Al2(SO4)3
d.
There are two moles of aluminum per mole of aluminum sulfate so we do:
2 x 0.00048 = 0.00096 moles of aluminum (rounded)
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