Answer to Question #158862 in General Chemistry for Syahidatul Hafizah Binti Rais

Question

Method: • 32.50 g of the impure magnesium sulfate is dissolved in water and the solution is made up to 500 mL in a volumetric flask. • Different volumes of 0.1 M barium chloride are added to six separate 20 mL samples of this solution. This precipitate formed as barium sulfate. • The precipitate from each sample is filtered, rinsed with deionised water and then dried in an oven until a constant mass of precipitate is obtained. Results: • Mass of impure magnesium sulfate: 32.50 g • Volume of volumetric flask: 500 mL • Volume of magnesium sulfate solution in each sample: 20 mL 

Calculate: i. The amount (in mole) of sulfate ion in the 500 mL volumetric flask. ii. The percent by mass of magnesium sulfate in the powder. iii. The percent by mass of sulfate in the powder.

Answer 1

MgSO₄ MM = 120.366g/mol

Mass = 32.50g

Moles $=\dfrac{mass}{MM}$

$=\dfrac{32.50g}{120.366g/mol}$

$=0.27mol$

Since mole ratio of Mg²⁺ : SO4^2- =1:1

Then moles of SO₄^2- = 0.27mol

Concentration of MgSO4

0.27 mol were in 500mL

? mol would be in 1000mL

$\dfrac{0.27mol x 1000mL}{500mL} = 0.54M$

Now total volume of MgSO4 that reacted with BaCl2 = 20mL x6 = 120mL

Moles of MgSO4 in 120mL of solution

$=\dfrac{0.54molx120mL}{1000mL} = 0.0648mol$

Moles of SO4 in this solution;

$=\dfrac{1}{1}x0.0648mol=0.0648mol$

NB: We require the mass of the precipitate formed to proceed from here