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  4. 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol-1 is :

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Q. 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 :

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Solution:

$M _{2} CO _{3}+2 HCl \rightarrow 2 MCl + H _{2} O + CO _{2}$

$n _{ M _{2} CO _{3}}= n _{ CO _{2}}$

$\frac{1}{ M _{ M _{2} CO _{3}}} =0.01186$

$M _{ M _{2} CO _{3}} =\frac{1}{0.01186}$

$= 84.3\, g / mol$