Question

Sol. [Key Concept: Number of atoms $=$ moles $\times$ atomicity $\times \ N_{A}$; computing each option, $2$ g $O_{2}$, $1400$ mL $O_{2}$ at STP and $0.0625$ mol $H_{2}$ all give $0.125\ N_{A}$.]

$N_{\text{atoms}\text{~}}$ in $O_{2} = \frac{m_{O_{2}}}{M_{O_{2}}}{\text{ }N}_{A} = \frac{2}{32} \times 2{\text{ }N}_{A} = \frac{N_{A}}{8} = 0.125{\text{ }N}_{A}$

$N_{\text{atoms}\text{~}}$ in ${SO}_{2} = \frac{m_{{SO}_{2}}}{M_{{SO}_{2}}} = 3{\text{ }N}_{A} = \frac{4}{64} \times 3{\text{ }N}_{A} = \frac{3{\text{ }N}_{A}}{16} = 0.1875{\text{ }N}_{A}$

$N_{\text{atoms}\text{~}}$ in $O_{2}$ at STP $= \frac{V_{O_{2}}}{22.7\text{~}\text{lit}\text{~}}N_{A} = \frac{1400ml}{22400ml} \times 2{\text{ }N}_{A} = 0.125{\text{ }N}_{A}$

$N_{\text{atoms}\text{~}}$ in $He(g)$ at $STP = \frac{V_{He}}{22.4\text{~}\text{lit.}\text{~}}N_{A} = \frac{0.05}{22.4}{\text{ }N}_{A} = 0.002{\text{ }N}_{A}$

$N_{\text{atoms}\text{~}}$ in $H_{2}(\text{ }g) =$ moles × atomicity $\times N_{A} = 0.0625 \times 2{\text{ }N}_{A} = 0.125{\text{ }N}_{A}$