Question

The density of air is $0.001293\,g/cc$ at $STP$ . Its vapour density is
(A) $0.001293$
(B) $8.2786$
(C) $14.48$
(D) $6.2706$

Answer 1

From the given density of the air at $STP$, calculate the molecular mass of the air. Since the molecular mass of the air is the mass filled in $22.4\,l$ of the air. From the calculated value, find the answer for the vapour density by using the formula.

Formula used:
(1) The vapour density is given by
$VD = \dfrac{{MM}}{2}$
Where $VD$ is the vapour density of the air and $MM$ is the molecular mass of the air.

Complete step by step solution:
It is given that the
Density of the air, $d = 0.001293\,gm{l^{ - 1}}$
Hence from the given data, it is clear that the $1\,ml$ of the air contains the $0.001293\,g$ of the molecules of the air. Then it is calculated for the $22.4\,l$ of the air. Since the $22.4\,l$ of the air gives the molecular mass of the air.
Molecular mass of the air is equal to the $22.4 \times {10^3}\,ml$ of the air.
$MM = 22.4 \times {10^3} \times 0.001293$
By simplifying the above equation,
$MM = 28.96\,g$
Hence the molecular mass of the air is calculated as $28.96\,g$ .
Using the formula of the vapour density,
$VD = \dfrac{{MM}}{2}$
Substituting the value of the molecular mass in the above formula.
$VD = \dfrac{{28.96}}{2}$
By dividing the terms in the right hand side of the equation, we get
$VD = 14.48$
Hence the vapour density of the given density of the air is calculated as $14.48$ .

Thus the option (B) is correct.

Note: At $STP$, which means standard temperature and the pressure of any compound, the mass at the $22.4$ litre is equal to the molecular or the molar mass. This is why the $22.4\,$ litre is substituted in one of the steps for the calculation of the molecular mass.