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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.To percentage of impurity in the sample is…........... |
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Answer» 0 `MgCO_(3):(1xx24)+(1xx12)+(3xx16)=84 g` `CO_(2):(1xx12)+(2xx16)=44 g` 100% pure 84 g `MgCO_(3)` on heating GIVES 44 g `CO_(2)` Given that 1 g of `MgCO_(3)` on heating gives `0.44 g CO_(2)` Therefore, 84 g `MgCO_(3)` sample on heating gives `36.96 g CO_(2)=84%` Percentage of purity of the sample =`(100%)/(44 g CO_(2))xx36.96 g CO_(2)=84%` Percentage of impurity=16% |
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