Question

The density of air is $0.001293\,{\text{g/}}\,{\text{mL}}$. Its Vapour density is:
A. ${\text{143}}$
B. ${\text{14}}{\text{.48}}$
C. ${\text{1}}{\text{.43}}$
D. $0.143$

Answer 1

Hint: Vapour density is determined by dividing the molecular mass of vapour by two. Molecular mass will be determined by using the density formula. The volume can be taken as the standard volume occupied by any gas at STP.

Formula used: ${\text{Vapour}}\,{\text{density}}\,{\text{ = }}\,\dfrac{{{\text{molecular}}\,{\text{mass}}}}{2}$

Complete step by step answer:
The density of vapour with respect to the hydrogen is defined as the Vapour density. Vapour density is unit less. At standard temperature and pressure, as all the gas have the same temperature and pressure so, the density becomes the function of molar mass only. The Vapour density and molecular mass is directly related at STP. The molecular mass is high means the gas has a high density or vice versa.
The formula to determine the Vapour density is as follows:
${\text{Vapour}}\,{\text{density}}\,{\text{ = }}\,\dfrac{{{\text{molecular}}\,{\text{mass}}}}{2}$
At STP one mole of a gas occupies $22.4\,{\text{L}}$ volume.
Convert the volume from L to mL as follows:
$1\,{\text{L}}\, = \,1000\,{\text{mL}}$
$\Rightarrow 22.4\,{\text{L}}\, = \,22400\,{\text{mL}}$
Determine the mass of the vapour as follows:
${\text{density}}\,{\text{ = }}\,\dfrac{{\,{\text{mass}}}}{{{\text{volume}}}}$
Substitute $22400\,{\text{mL/mol}}$for volume and $0.001293\,{\text{g/}}\,{\text{mL}}$ for density.
$\Rightarrow 0.001293\,{\text{g/}}\,{\text{mL}}\,{\text{ = }}\,\dfrac{{\,{\text{mass}}}}{{22400\,{\text{mL/mol}}}}$
$\Rightarrow {\text{mass}}\, = \,0.001293\,{\text{g/}}\,{\text{mL}}\,\, \times \,22400\,{\text{mL/mol}}$
$\Rightarrow {\text{mass}}\, = \,28.96\,{\text{g/mol}}$
So, the mass of vapour is $\,28.96\,{\text{g/mol}}$
Determine the Vapour density is as follows:
Substitute $28.96$ for molecular mass
$\Rightarrow {\text{Vapour}}\,{\text{density}}\,{\text{ = }}\,\dfrac{{\,28.96}}{2}$
$\Rightarrow {\text{Vapour}}\,{\text{density}}\,{\text{ = }}\,14.48\,$
So, the Vapour density is $14.48$.

Therefore, option (B) $14.48$ is correct.

Note:
STP is known as standard temperature and pressure. The value of standard temperature is $273\,{\text{K}}$. The value of standard pressure is ${\text{1}}\,{\text{atm}}$. One mole of a gas at $273\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$ volume. The Vapour density is used to determine whether a gas is more or less dense than the air.