1 g of a carbonate $(\mathrm{M}_{2}\mathrm{CO}_{3})$ on treatment with excess HCl produces 0.01 mol of $\mathrm{CO}_{2}$. The molar mass of $\mathrm{M}_{2}\mathrm{CO}_{3}$ is ______ $\mathrm{g~mol}^{-1}$. (Nearest integer)

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Some Basic Concepts of Chemistry

1 g of a carbonate

(M_{2} CO_{3})

on treatment with excess HCl produces 0.01 mol of

CO_{2}

. The molar mass of

M_{2} CO_{3}

is ______

g mol^{- 1}

. (Nearest integer)

Answer: 100

The reaction is:

M_{2} CO_{3} + 2 HCl \to 2 MCl + H_{2} O + CO_{2}

. By atomic conservation of carbon: moles of

M_{2} CO_{3}

= moles of

CO_{2}

= 0.01 mol. Since 1 g of

M_{2} CO_{3}

gives 0.01 mol, molar mass =

\frac{1 g}{0.01 mol} = 100 g mol^{- 1}

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer