Match List-I with List-II.

	List - I		List - II
a)	4.48 litres of O2 at STP	I)	0.2 moles
b)	12.022×1022 molecules of H2O	II)	12.044×1023 molecules
c)	96g of O2	III)	6.4g
d)	88 g CO2	IV)	67.2 litres at STP

(Given-Molar volume of a gas at STP =22.4 L )
Choose the correct answer from the options given below.

(a) 4.48 litres of O₂ at STP →6.4g
$\begin{array}{l}{\mathrm{n}}_{{\mathrm{O}}_2}=\frac{\mathrm{V}}{22.4}=\frac{4.48}{22.4}=0.2\text{ moles }\\ \mathrm{w}=\mathrm{n}\times {\mathrm{M}}_{{\mathrm{O}}_2}=0.2\times 32=6.4\mathrm{g}\end{array}$
(b) 12.022×10²² molecules of H₂O→0.2 moles
${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}=\frac{12.022\times {10}^{22}}{6.022\times {10}^{23}}=2\times {10}^{-1}=0.2\text{ moles }$
c) 96g of O₂→67.2 litres at STP
$\begin{array}{l}\frac{\mathrm{w}}{\mathrm{M}}=\frac{\mathrm{V}}{22.4}\\ \frac{96}{32}=\frac{\mathrm{V}}{22.4}\mathrm{V}=\frac{96\times 22.4}{32}=67.2\text{ litres }\end{array}$
(d) 88g of CO₂→12.044x10²³ molecules
$\begin{array}{l}\frac{\mathrm{W}}{\mathrm{M}}=\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}\therefore \frac{88}{44}=\frac{\mathrm{N}}{6.022\times {10}^{23}}\\ \mathrm{N}=6.022\times {10}^{23}\times 2=12.044\times {10}^{23}\text{ molecules }\end{array}$