The molar mass \(\left( H _{2}\right) 2 \times 1=2 g / mol\)
No of mole of \(H _{2}\)
\begin{aligned}
&=\frac{\text { mass of } H _{2}}{ m . \text { mass of } H _{2}} \\
&=\frac{4.00 g }{2.000 g / mol }=2.0 mol \\
\text { From } P V=n R T \\
V &=\frac{n R T}{P} \\
&=\frac{2.0 \times 0.082 \times(27+273)}{2} \\
&=24.6 \text { litre }
\end{aligned}
\(\therefore\) its volume is \(24.6\) litre.ut,
No of \(mol =\frac{\text { no of molecules }}{6.02 \times 10^{23}}\)
\(\therefore\) no of molecule
= no of mole \(6.02 \times 10^{23}\)
\(=2.0 mol \times 6.02 \times 10^{23}=12.04 \times 10^{23}\)
i.e. it contains \(12.04 \times 10^{23}\) molecules and not \(6.02 \times 10^{23}\) molecules