Question

a) How many moles of Al2(SO4)3 are required to make 43 mL of a 0.060 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(S04)3, to the nearest gram? grams c) What mass of Al2(SO4)3 is required to make 43 mL of a 0.060 M Al2(Soa)3 solution? (If you've got moles and molecular weight you can get to grams.) g Al2(SO4)3 d) How many moles of sulfate 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 sulfate ions there are per Al2(SO4)3-) mol SO2-

Answer 1

1)
volume , V = 43 mL
= 0.43 L


use:
number of mol,
n = Molarity * Volume
= 0.060*0.43
= 2.6*10^-3 mol
Answer: 2.6*10^-3 mol

2)
Molar mass of Al2(SO4)3,
MM = 2*MM(Al) + 3*MM(S) + 12*MM(O)
= 2*26.98 + 3*32.07 + 12*16.0
= 342.17 g/mol
Answer: 342.17 g/mol

3)
use:
mass of Al2(SO4)3,
m = number of mol * molar mass
= 2.6*10^-3 mol * 342.17 g/mol
= 0.88 g
Answer: 0.88 g

4)
1 mol of Al2(SO4)3 has 3 mol of SO42- ions
So,
Mol of SO42- = 3*number of mol of Al2(SO4)3
= 3*2.6*10^-3 mol
= 7.8*10^-3 mol
Answer: 7.8*10^-3 mol