1 gram of a carbonate $(\mathrm{M}_{2} \mathrm{CO}_{3})$ on treatment with excess HCl produces 0.01186 mole of $\mathrm{CO}_{2}$. The molar mass of $\mathrm{M}_{2} \mathrm{CO}_{3}$ in $\mathrm{g} \mathrm{~mol}^{-1}$ is :

JEE Main · 2017 · 02 Apr 2017 · Offline · Some Basic Concepts of Chemistry

1 gram of a carbonate

(M_{2} CO_{3})

on treatment with excess HCl produces 0.01186 mole of

CO_{2}

. The molar mass of

M_{2} CO_{3}

g mol^{- 1}

is :

Answer: (B)

One mole of

M X_{2} CO X_{3}

releases one mole of

CO X_{2}

on treatment with acid, so the sample contains

0.01186

mol of carbonate. Its molar mass is then

\frac{1 g}{0.01186 mol} = 84.3

g mol

^{- 1}

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option