Question

$10g$ of $MgC{O}_3$ decomposes on heating to $0.1mole$ $C{O}_2$ and $4gMgO$. The percent purity of $MgC{O}_3$ is

• A. 24%
• B. 44%
• C. 54%
• D. 74%
• E. 84%

Answer 1

Answer: (E)

Since, $MgC{O}_3\stackrel{\Delta }{\to }MgO+C{O}_{2(g)}$
Mole ratio $1:1:1$
Mass ratio (ing) $84:40:44$
Given values $10g4g0.1mol$
$(=4.4g)\left\{\because n=\frac{W}{m}\right\}$
(i) $\because 84g.$ of $MgC{O}_3$ give $MgO=40g$
$\therefore 10g$ of $MgC{O}_3$ give $MgO=\frac{40\times 10}{84}=4.76g$
Thus, $MgO$ obtained is less by
$4.76-4.0=0.76g$
(ii) $\because 84g$ of $MgC{O}_3$ give $C{O}_2=44g$
$\therefore 10g$ of $MgC{O}_3$ give $C{O}_2=\frac{44\times 10}{84}=5.23g$
Thus $C{O}_2$ obtained is less by $5\cdot 23-4\cdot 4=0\cdot 84g$
Total short $=0\cdot 76+0\cdot 84=1\cdot 59=1.60$
$\because 10g.$ of $MgC{O}_3$ give less products by $=1.60g$
$\therefore 100g$ of $MgC{O}_3$ give less products $by=1.600g$
Hence, $MgC{O}_3=100-16=84\%$