Calculate the number of moles, molecules and total number of atoms present in $5.6c{m}^3$ of $N{H}_3$ at STP.
[Give $C=12,O=16,N=14,H=1,He=4$]

Volume of $N{H}_3$ at STP= $5.6c{m}^3⟶5.6\times {10}^{-3}d{m}^3$

(i) Moles of $N{H}_3=\frac{VolumeofgasatSTP}{22.4d{m}^3}=\frac{5.6\times {10}^{-3}d{m}^3}{22.4d{m}^3}=2.5\times {10}^{-4}mol$

(ii) Number of molecules= $moles\times Avogadr{o}^{\prime }snumber$

$=2.5\times {10}^{-4}\times 6.022\times {10}^{23}$

$=1.5\times {10}^{20}$ molecules

(iii) Ammonia has total $4$ atom

$\therefore$ Number of atoms in $5.6\times {10}^{-3}d{m}^3$

$=1.5\times {10}^{20}\times 4$

$=6\times {10}^{20}$ atoms