Question

In: Chemistry

Aluminum sulphate is used in the manufacture of paper and in the water purification industry. In...

Aluminum sulphate is used in the manufacture of paper and in the water purification industry. In the solid state, aluminum sulphate is a hydrate. The formula is Al2(SO4)3.18H2O. How many grams of sulphur are there in 0.479 moles of solid aluminum sulphate?

How many water molecules are there in a 4.63 g sample of solid aluminum sulphate?

How many moles of sulphate ions are there in a sample of solid aluminum sulphate that contains 13.7 moles of oxygen atoms?

An aqueous solution of aluminum sulphate contains 1.46 % by mass aluminum and has a density of 1.05 g/mL. What is the concentration of aluminum ions in the solution?

Solutions

Expert Solution

The formula is Al2(SO4)3.18H2O.

1) How many grams of sulphur are there in 0.479 moles of solid aluminum sulphate?

moles of S in aluminum sulphate = 3

moles of S in 0.479 mol aluminum sulphate = 3 x 0.479

                                                                 = 1.437 mol

mass of sulphur = 1.437 x 32 = 45.98 g

mass of sulphur = 46.0 g

2) How many water molecules are there in a 4.63 g sample of solid aluminum sulphate?

mass of aluminum sulphate = 4.63 g

moles of aluminum sulphate = 4.63 / 666.4 = 0.00695 mol

1 mol Al2(SO4)3.18H2O   --------------> 18 mol

0.00695 mol Al2(SO4)3.18H2O   -------------> ??

moles of water = 0.00695 x 18

                             = 0.125

number of water molecules = 0.125 x 6.023 x 10^23

                                                = 7.53 x 10^22

3)

How many moles of sulphate ions are there in a sample of solid aluminum sulphate that contains 13.7 moles of oxygen atoms?

1 mol Al2(SO4)3.18H2O.   ----------> 30 mol O atoms

      ?? Al2(SO4)3.18H2O.    ----------> 13.7 mol O atoms

moles of Al2(SO4)3.18H2O. = 13.7 x 1 / 30 = 0.457 mol

number of moles of sulphate ions = 0.457 x 3 = 1.37 mol

number of moles of sulphate ions = 1.37 mol

4)

Concentration = 1.46 x 10 x 1.05 / 27

                        = 0.568 M


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