Question

In the decomposition of 10 g of $Mg{CO}_3$, 0.1 mole ${CO}_2$ and 4.0 g MgO are obtained. Hence, percentage purity of $Mg{CO}_3$ is:

• A. 50%
• B. 60%
• C. 40%
• D. 84%

Answer 1

Answer: (D)

84%

Solution:- (D) $84\%$

Molecular weight of $MgC{O}_3=84g$

Molecular weight of $MgO=40g$

Decomposition of $MgC{O}_3$-

$MgC{O}_3⟶MgO+C{O}_2$

Now, from the above reaction-

Weight of pure $MgC{O}_3$ required to produce $40g$ of $MgO=84.3g$

Weight of pure $MgC{O}_3$ required to produce $4g$ of $MgO=\frac{84.3}{40}\times 4=8.43g$

Given weight of $MgC{O}_3=10g$

Now,

Amount of pure $MgC{O}_3$ in $10g$ of given $MgC{O}_3=8.43g$

Thus,

Amount of pure $MgC{O}_3$ in $100g$ of given $MgC{O}_3=\frac{8.43}{10}\times 100=84.3g$

Therefore,

The percentage purity of given $MgC{O}_3=84.3\%\approx 84\%$