Question

: Vapour Density of a gas is \[11.2\]. The volume occupied by \[11.2\] g of this gas at STP is:
A.$22.41$
B.$11.21$
C.$11$
D.$2.251$

Answer 1

Hint: The question can be solved from the knowledge of the relation between vapour density and molecular weight of a gas. Substitute the values in the equation given below to find the molar mass of gas and thus, the volume.

Formula used:
${\text{V}}{\text{.D }} = \dfrac{{{\text{molar mass of gas}}}}{{{\text{molar mass of hydrogen}}}}$

Complete step by step answer:
The vapour density of a gas is the density of the gas in relation to hydrogen. Alternatively, it can be defined as the mass of a certain volume of a gas by the mass of the same volume of hydrogen.
Mathematically, Vapour Density = $\dfrac{{{\text{Molar mas}}s}}{2}$
Putting the value of the vapour density of the gas in the above equation we get:
Molar Mass of the gas$ = 2 \times 11.2$= $22.4$ ${\text{g/mol}}$
Now it is known that for all ideal gases, the molar mass of the gas = $22.4$litre under standard conditions of temperature and pressure.
If $22.4$ ${\text{g/mol}}$= $22.4$ litre at STP, then
\[11.2\] g of this gas = \[11.2\] litre at STP.

Hence, the correct answer is option B.

Note:
The vapour density of the gas is equal to the ratio of the mass of n molecules of the gas to the mass of n molecules of hydrogen. Therefore, it is also the ratio of the molar mass of the gas to the molar mass of hydrogen. Hence the formula of vapour density is obtained.
The statement that a gas has vapour density 2 in relation to air means that the gas is twice as heavy as air. The vapour density of a gas indicates how many times a gas is heavier than air.