Question

a) How many moles of Al2(SO4)3 are required to make 44 mL of a 0.090 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 44 mL of a 0.090 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 A13+

Answer 1

a) moles = molarity * volume in liters

moles of Al₂(SO₄)₃ = 0.090 M * 0.044 L => 0.00396 moles

b) Al₂(SO₄)₃ molecular weight => [ (2*Al) + (3*s) + (12*O) ] => [ (2*26.98) + (3*32.065) + (12*16) ] => 342.15 grams

c) mass of Al₂(SO₄)₃ => 0.00396 mol * 342.15 g/mol => 1.355 grams

d) 1 mol Al₂(SO₄)₃ contains the 2 moles of Al3+

1mol Al₂(SO₄)₃ --------------------> 2 mol Al3+

0.00396 mol ---------------------->?

=> 0.00396 * 2 => 0.00792 moles