Question

1 gram of a carbonate $(M_2CO_3)$ on treatment with excess $HCl$ produces $0.01186$ mole of $CO_2$. The molar mass of $M_2CO_3$ in g $mol^{-1}$ is :

• A. $118.6$
• B. $11.86$
• C. $1186$
• D. $84.3$

Answer 1

Answer: (D)

$84.3$

Answer 2

$M _{2} CO _{3}+2 HCl \rightarrow 2 MCl + H _{2} O + CO _{2}$
$n _{ M _{2} CO _{3}}= n _{ CO _{2}}$
$\frac{1}{ M _{ M _{2} CO _{3}}} =0.01186$
$M _{ M _{2} CO _{3}} =\frac{1}{0.01186}$
$= 84.3\, g / mol$