Question

$\mathrm{Calculate}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{molecules}\mathrm{present}\mathrm{in}700{\mathrm{cm}}^3{\mathrm{NH}}_3\mathrm{at}\mathrm{S}.\mathrm{T}.\mathrm{P}.$

Answer 1

$$\begin{gathered} 1\mathrm{mole}\mathrm{of}{\mathrm{NH}}_3=6.022\times {10}^{23}\mathrm{molecules}=22400{\mathrm{cm}}^3\mathrm{at}\mathrm{S}.\mathrm{T}.\mathrm{P}. \\ \mathrm{Thus},22400{\mathrm{cm}}^3\mathrm{of}{\mathrm{NH}}_3\mathrm{at}\mathrm{S}.\mathrm{T}.\mathrm{P}.\mathrm{contains}=6.022\times {10}^{23}\mathrm{molecules} \\ \mathrm{Therefore},700{\mathrm{cm}}^3\mathrm{of}{\mathrm{NH}}_3\mathrm{at}\mathrm{S}.\mathrm{T}.\mathrm{P}.\mathrm{contains}=\frac{6.022\times {10}^{23}\times 700}{22400} \\ =1.882\times {10}^{22}\mathrm{molecules} \\ \mathrm{Therefore},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{molecules}\mathrm{present}\mathrm{in}700{\mathrm{cm}}^3\mathrm{of}{\mathrm{NH}}_3\mathrm{at}\mathrm{S}.\mathrm{T}.\mathrm{P}.\mathrm{is}1.882\times {10}^{22}\mathrm{molecules}. \end{gathered}$$