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

A. 0
B. 0.044
C. 0.16
D. 0.084

Answer

C

Explanation

`MgCO_(3) to MgO +CO_(2) uarr`
`MgCO_(3): (1xx24) + (1xx12) + (3xx16)=84g`
100% pure 84 g of `MgCO_(3)` on heating gives 44g `CO_(2)`
Given that 1g of `MgCO_(3)` on heating gives 0.44g `CO_(2)`
Therefore, 854 g `MgCO_(3)` sample on heating gives 36.96 g `CO_(2)`
Percantaqge of purity of the sample= `(100%)/(44 g CO_(2)) xx36.96g CO_(2)`
=84%
Percentage of impurity=16%