$113.5 \mathrm{mL}$of ${\mathrm{NO}}_2$ at STP was liquefied, the density of the liquid being $1.15 \mathrm{g}$ mL1. Calculate the volume of ${\mathrm{NO}}_2$ and the number of molecules in the liquid ${\mathrm{NO}}_2$. (At. Wt. N = 14)

Calculate the volume of ${\mathrm{NO}}_2$

As we know,

$\begin{array}{l}{n}_{{\mathrm{NO}}_2}=\frac{113.5}{22.7}=5\times {10}^{-3},m_{{\mathrm{NO}}_2}=5\times 46\times {10}^{-3}\\ \begin{array}{rl}{V}_{{\mathrm{NO}}_2\left(1\right)}& =\frac{5\times 46\times {10}^{-3}}{1.15}=0.200\mathrm{mL}\\ N& =5\times {10}^{-3}\times 6.02\times {10}^{23}\\ & =3.01\times {10}^{21}\end{array}\end{array}$

Thus, the volume of ${\mathrm{NO}}_2$ and the number of molecules in the liquid  $0.20\mathrm{mL}$ and $3.01\times {10}^{21}$.