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…...........

1 g of an impure sample of magnesium carbonate (containing no thermall

1.	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…...........
Answer» <html><body><p>0<br/>`4.4%`<br/>0.16<br/>`8.4%`</p>Solution :`MgCO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)toMgO+CO_(2)<a href="https://interviewquestions.tuteehub.com/tag/uarr-3241817" style="font-weight:bold;" target="_blank" title="Click to know more about UARR">UARR</a>`<br/> `MgCO_(3):(1xx24)+(1xx12)+(3xx16)=84 g`<br/> `CO_(2):(1xx12)+(2xx16)=44 g`<br/> 100% pure 84 g `MgCO_(3)` on heating <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> 44 g `CO_(2)`<br/> Given that <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> g of `MgCO_(3)` on heating gives `0.44 g CO_(2)`<br/> Therefore, 84 g `MgCO_(3)` sample on heating gives `36.96 g CO_(2)=84%`<br/> Percentage of purity of the sample =`(100%)/(44 g CO_(2))xx36.96 g CO_(2)=84%`<br/> Percentage of impurity=16%</body></html>