$$0.24\,g$$  of a volatile liquid on vaporization gives $$45\,ml$$  of vapours at $$NTP.$$  What will be the vapour density of the substance ?
$$\left( {{\text{Density of}}\,{\text{ }}{H_2} = 0.089{\text{ }}g{\text{ }}{L^{ - 1}}} \right)$$

Solution :
$$\eqalign{ & V.D. = \frac{{Wt.\,{\text{of}}\,\,45\,ml.\,{\text{of}}\,\,{\text{vapours at }}NTP}}{{Wt.\,{\text{of}}\,\,45\,ml.\,{\text{of}}\,{H_2}\,\,{\text{at }}NTP}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.24\,g}}{{45ml. \times 0.000089\,g\,m{l^{ - 1}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 59.93 \cr} $$