Number of atoms present in 1.6 grams of methane are$\left({\mathrm{N}}_{\mathrm{A}}=\mathrm{Avagadro}\text{'}\mathrm{s} \mathrm{number}\right)$

$$\begin{gathered} 16 \mathrm{g} {\mathrm{CH}}_{4 }\mathrm{contain} {\mathrm{N}}_{\mathrm{A}} \mathrm{molecules} \mathrm{and} 5{\mathrm{N}}_{\mathrm{A}} \mathrm{atoms}; \\ 1.6 \mathrm{g} {\mathrm{CH}}_{4 }\mathrm{contain} 0.1{\mathrm{N}}_{\mathrm{A}} \mathrm{molecules} \mathrm{and} 0.5{\mathrm{N}}_{\mathrm{A}} \mathrm{atoms}; \end{gathered}$$