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(B) According to Avogadro's Law,at the same temperature and pressure,equal volumes of gases contain an equal number of moles.Since the volume of the v

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$\frac{1}{2}$ that of $SO_2$

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(B) According to Avogadro's Law,at the same temperature and pressure,equal volumes of gases contain an equal number of moles.Since the volume of the vessel is constant,the number of moles of $O_2$ and $SO_2$ will be the same.Weight of gas = $	ext{Number of moles} 	imes 	ext{Molar mass}$.Since the number of moles is the same,the weight of the gas is directly proportional to its molar mass.Weight of $O_2$ / Weight of $SO_2$ = $\frac{	ext{Molar mass of } O_2}{	ext{Molar mass of } SO_2} = \frac{32}{64} = \frac{1}{2}$.Therefore,the weight of oxygen will be $\frac{1}{2}$ that of $SO_2$.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

$A$ pre-weighed vessel was filled with oxygen at $N.T.P.$ and weighed. It was then evacuated,filled with $SO_2$ at the same temperature and pressure,and again weighed. The weight of oxygen will be

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