Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85 g of BaCO₃ (Atomic mass of Ba = 137) will be

${\mathrm{BaCO}}_3\to \mathrm{BaO}+{\mathrm{CO}}_2\uparrow$
Molecular weight of BaCO₃ = 137 +12 + 3 x 16 = 197
$\because$$197 \mathrm{gm} \mathrm{produces} 22.4 \mathrm{L} \mathrm{at} \mathrm{S}.\mathrm{T}.\mathrm{P}.$
$\therefore 9.85 \mathrm{gm} \mathrm{produces} \frac{22.4}{197}\mathrm{x} 9.85 = 1.12 \mathrm{L} \mathrm{at} \mathrm{S}.\mathrm{T}.\mathrm{P}.$