Question

A 0.789g sample of impure magnesium oxide was dissolved in 25 cm3 of 1 mol dm-3 Sulfuric acid. The resulting solution was made up to 100cm3 with water. 25 cm3 of this final solution were neutralized by 27.8 cm3 of 0.1 mol dm-3 sodium hydroxide. Calculate the percentage purity of the magnesium oxide.

Answer 1

0.789g MgO = 0.019725 mol -

concentration is final solution:

c = 0.0019725 M and 0.0042775M of acid

This solution was neutralised by:

V*C= 0.0278*0.1*4 = 0.0112 mol

0.025-0.019725x= 0.0112 mol

Purity is 69.9%