20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. wt. of Mg = 24)
- 96
- 60
- 84
- 75
Answer
(3) 84
Solution
MgCO3 MgO + CO2
Molar mass of MgCO3 = 24 + 12 + 48 = 84 g mol-1; molar mass of MgO = 24 + 16 = 40 g mol-1
⇒ Moles of MgO formed = 8.040 = 0.2 mol
⇒ Moles of MgCO3 decomposed = 0.2 mol
⇒ Mass of pure MgCO3 = 0.2 84 = 16.8 g
∴ Percentage purity = 16.820.0 100 = 84\%
Answer: (3)