Question

a) How many moles of Al₂(SO₄)₃ are required to make 55 mL of a 0.050 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 55 mL of a 0.050 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⁺

Answer 1

Melaidy ser af aeles ofsaluls Valurne of salutaon males Malarly Volume O.05M X o.055 L O-00275 males 55 ml (65/1000) L =0.055L Makiular weyht of Al oy)-a aternie massfa +3(atomie mass of ) +12 (atomue mas 2 X 271/mel +3 x 32 9lma +12 x 16 g/mal 342g mal melardy mass Melarmass X Valume Malardy xMalar mau xValume 005 mel/L X 342 glmal x 0-055L 0- 44053 mass = O.9 Nato!meles=mass Molay'mas + 2AL +3s0, 2 X meles of Al, s,] 2 X000275 meles males ef Al O-0055males 1