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Question
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is
Options
0%
4.4%
16%
8.4%
Solution
16%
Explanation:
\[\ce{MgCO3 -> MgO + CO2 ^}\]
MgCO3: (1 × 24) + (1 × 12) + (3 × 16) = 84 g
CO2: (1 × 12) + (2 × 16) = 44g
100% pure 84 g MgCO3 on heating gives 44 g CO2
Given that 1 g of MgCO3 on heating gives 0.44 g CO2
Therefore, 84 g MgCO3 sample on heating gives 36.96 g CO2 = 100%
Percentage of purity of the sample = `(100%)/(44 "g CO"_2) xx 36.96` g CO2
= 84%
Percentage of impurity = 16%
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