Rearrange the following (I to IV) in the orde | vedclass

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(A) To find the order of increasing masses,we calculate the mass of each component:$(I)$ Mass of $0.5$ mole of $O_3 = 0.5 	imes 48 \, g = 24 \, g$.$(

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$II < IV < III < I$

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(A) To find the order of increasing masses,we calculate the mass of each component:$(I)$ Mass of $0.5$ mole of $O_3 = 0.5 	imes 48 \, g = 24 \, g$.$(II)$ $0.5 \, g$ atom of oxygen means $0.5$ mole of $O$ atoms. Mass $= 0.5 	imes 16 \, g = 8 \, g$.$(III)$ Number of moles of $O_2 = \frac{3.011 	imes 10^{23}}{6.022 	imes 10^{23}} = 0.5$ mole. Mass $= 0.5 	imes 32 \, g = 16 \, g$.$(IV)$ Number of moles of $CO_2 = \frac{5.6 \, L}{22.4 \, L/mol} = 0.25$ mole. Mass $= 0.25 	imes 44 \, g = 11 \, g$.Comparing the masses: $8 \, g (II) Thus,the correct order is $II < IV < III < I$.

List-$I$	List-$II$
$(a)$ $2 \text{ moles of ethene}$	$(i)$ $11.2 \text{ L volume at STP}$
$(b)$ $\text{Molar mass is equal to } 66 \text{ g}$	$(ii)$ $56 \text{ g}$
$(c)$ $1 \text{ g of } H_2$	$(iii)$ $12.04 \times 10^{23} \text{ molecules}$
$(d)$ $2 \text{ moles of water vapours}$	$(iv)$ $1.5 \text{ mole of } CO_2$

Rearrange the following ($I$ to $IV$) in the order of increasing masses:
$(I)$ $0.5$ mole of $O_3$
$(II)$ $0.5 \, g$ atom of oxygen
$(III)$ $3.011 \times 10^{23}$ molecules of $O_2$
$(IV)$ $5.6 \, L$ of $CO_2$ at $STP$

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Rearrange the following ($I$ to $IV$) in the order of increasing masses:$(I)$ $0.5$ mole of $O_3$$(II)$ $0.5 \, g$ atom of oxygen$(III)$ $3.011 \times 10^{23}$ molecules of $O_2$$(IV)$ $5.6 \, L$ of $CO_2$ at $STP$