($2$) $4 \mathrm{~g}$ of Helium $=\frac{4 \mathrm{~g}}{4 \mathrm{~g}}$ mole $=1 \mathrm{~mole}=\mathrm{NA}_{\mathrm{Ae}}$ atom
($3$) $2.2710982 \text { of He at STP } $$ =\frac{2.271}{22.710982} \text { mole } $
$ =0.1 \mathrm{~mole} $
$ =0.1 \mathrm{~N}_{\mathrm{A}} \mathrm{He} \text { atom }$
($4$) $4 \mathrm{~mol}$ of $\mathrm{He}=4 \mathrm{~N} \mathrm{Ne}$ atoms
$Na _2 O + H _2 O \rightarrow 2 X$
$Cl _2 O _7+ H _2 O \rightarrow 2 Y$