$$0.24 g$$  of a volatile gas, upon vaporisation, gives $$45 mL$$  vapour at $$NTP.$$  What will be the vapour density of the substance?
$$\left( {{\text{Density of}}\,{H_2} = 0.089} \right)$$

Solution :
$$\eqalign{ & {\text{Weight of gas}} = 0.24\,g \cr & {\text{Volume of gas}}\left( V \right) = 45\,mL = 0.045\,L \cr & {\text{Density of}}\,{H_2}\left( d \right) = 0.089 \cr & {\text{Weight of }}45{\text{ }}mL{\text{ of }}{H_2} = V \times d \cr & = 0.045 \times 0.089 \cr & = 4.005 \times {10^{ - 3}}g \cr & {\text{Therefore, vapour density}} \cr & = \frac{{{\text{Weight of certain volume of substance}}}}{{{\text{Weight of same volume of hydrogen}}}} \cr & = \frac{{0.24}}{{4.005 \times {{10}^{ - 3}}}} \cr & = 59.93 \cr} $$