Question

The vapour density of gas is $22$. It cannot be:
A.Carbon dioxide
B.Nitrous oxide
C.Propane
D.Methane

Answer 1

Vapour density: The vapour phase is defined as the relative weight of vapour or gas in comparison to atmospheric air if the vapour density of any gas is higher than that of atmospheric air then it will rise above air, if vapour density of the gas is less than atmospheric air then it will be at lower level than atmospheric air.

Complete step by step answer:
In order to find out the gas present, we will study about the atomic weight of the gas.
Atomic weight: It is defined as the average atomic mass of the chemical element and is calculated as the sum total of the elements involved in the molecular structure.
Now we will calculate the atomic weight of each gas and will substitute in the relation involving both molecular weight and vapour density I.e.; $\Rightarrow Molecular{\text{ }}weight{\text{ }} = {\text{ }}2 \times vapour{\text{ }}density{\text{ }}of{\text{ }}the{\text{ }}gas........(1)$
Molecular weight of carbon dioxide:
$\Rightarrow C{O_2} = {\text{ mass of carbon + mass of 2 oxygen}} \\\ \Rightarrow C{O_2} = {\text{ 12 + 32}} \\\ \Rightarrow C{O_2} = {\text{ 44}} \\\$
Molecular weight of nitrous oxide:

$$\Rightarrow {N_2}O = {\text{ mass of 2 nitrogen + mass of oxygen}} \\\ \Rightarrow {N_2}O = {\text{ 28 + 16}} \\\ \Rightarrow {N_2}O = {\text{ 44}} \\\$$

Molecular weight of propane:

$$\Rightarrow {C_3}{H_8} = mass{\text{ }}of{\text{ }}3{\text{ }}carbon{\text{ }} + {\text{ }}mass{\text{ }}of{\text{ }}8{\text{ }}hydrogen \\\ \Rightarrow {C_3}{H_8} = 3 \times 12 + {\text{ }}1 \times 8 \\\ \Rightarrow {C_3}H{\text{ }} = {\text{ }}36 + 8 \\\ \therefore {C_3}H{\text{ }} = {\text{ }}44 \\\$$

Molecular weight of methane:
$\Rightarrow C{H_3} = {\text{ mass of carbon + mass of 3 hydrogen}} \\\ \Rightarrow C{H_3}{\text{ = 12 + 3}} \\\ \therefore C{H_3} = {\text{ 15}} \\\$
According to the equation (1), the molecular weight should be 44 as the vapour density given is $\;22$.
Atomic mass of carbon dioxide calculated from the formula in equation (1) is: $\;44$
Atomic mass of nitrous oxide calculated from the formula in equation (1) is $\;44$
Atomic mass of propane calculated from the formula in equation (1) is $\;44$
Atomic mass of methane calculated from the formula in equation (1) is $\;15$
Therefore the correct option is D i.e., methane since its atomic number is $\;15$ and not $\;44$.

Note:
Density and vapour density are two different aspects in order to define, density is the ratio of mass of substance to its volume whereas vapour density is defined as the ratio of the volume of certain gas to the equal volume of hydrogen gas.