Question

0.308 g magnesiumsulphate, that contains crystal water, is dissolved in water, så the total volume is 125.0 mL. To completely precipitate the suflation as barium sulfate in 25.0 mL of this solution, 5.0 mL of 0.050 M barium hydroxide is required.

Calculate how many moles of crystal water are in 1 mole of magnesium sulfate.

I am lost in this one, hope for some help :)

Answer 1

Let the magnesium sulphate contains x molecules of H₂O attached with it.

Therefore the molecular formula will be MgSO₄.xH₂O

Therefore the molecular mass = 24.32 + 32 + 64 + 18x = [120.32 + 18x] g mol^-1

The amount of MgSO₄.xH₂O taken is 0.308 g = 0.308 / [120.32 + 18x] moles.of MgSO4.xH2O

Since 1 mole of MgSO4.xH2O contains 1 mole of MgSO4, it is equal to moles of MgSO4

This much moles is dissolved and diluted to 125 mL and 25 mL is taken for precipitation as BaSO₄

Therefore the 25 mL of the alquot contains [25/125] * 0.308 / [ 120.32 + 18x] moles of MgSO4

                                                            = 0.0616 / [120.32 + 18x] moles of MgSO4

In the precipitation reaction MgSO4 + Ba(OH)2 BaSO4 + Mg(OH)2

1 mole of MgSO4. contains one mole of SO4 ions, the equivalent number of moles of Ba²⁺ required is also equal to the number of moles of MgSO4.

Therefore the number of moles of Ba²⁺ ions required is = 0.0616 / [120.32 + 18x] moles ------------------(1)

Given that it consumes 5 ml of 0.05 M Ba(OH)₂

5 mL of 0.05 M solution = 5 * 0.05 / 1000 moles of Ba²⁺ ions = 0.00025 moles. ----------------- (2)

The number of moles of Ba²⁺ ions required in (1) = the number of moles of Ba²⁺ ions consumed in (2)

Therefore 0.0616 / [120.32 + 18x] = 0.00025

             0.0616 = 0.03008 + 0.0045x

.            0.03152 = 0.0045x

            x = 0.03152 / 0.0045 = 315.2 / 45 = 7

Therefore the molecular formula of the given substance is MgSO₄.7H₂O

Hence there are 7 moles of crystal water in one mole of magnesium sulphate.