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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 …………(a) 0% (b) 4.4% (c) 16% (d) 8.4% |
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Answer» Correct answer is (c) 16% MgCO3 → MgO + CO2 ↑ MgCO3 : (1 × 24) + (1 × 12) + (3 × 16) = 84g 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 Percentage of purity of the sample = \(\frac{100}{44gCO_2}\) x 36.96 g CO2 = 84% Percentage of impurity = 16% |
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