Question

# Vapour density of a gas is 11.2. The volume occupied by 11.2 g of this gas at STP is:(A) 22.4 L(B) 11.2 L(C) 1 L(D) 2.25 L

Hint: Do not confuse between vapour density and density. Vapour density is the density of a vapour in relation to that of a particular gas at the same volume. It is a ratio of molar masses.

Complete step by step answer:
Vapour density of a gas is equal to the ratio of Molar mass of some volume of a certain gas to Molar mass of the same volume of hydrogen gas - ${{H}_{2}}$.
Molar mass of hydrogen gas = 2 x Molar mass of hydrogen = 2 x 1 u = 2 u.
Therefore,
Vapour density $=\dfrac{Molar\text{ }mass\text{ }of\text{ }gas}{2}$
(Given, vapour density = 11.2)
11.2 $=\dfrac{Molar\text{ }mass\text{ }of\text{ }gas}{2}$
So, Molar mass of gas = 22.4
Now, let’s calculate the number of moles of gas
(Given, weight of gas = 11.2 g)
$Moles=\dfrac{Weight}{Molar\text{ }Mass}=\dfrac{11.2}{22.4}=\text{0}\text{.5 }moles$
Avogadro’s hypothesis states that at standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Therefore, we can write the following –
At STP, 1 mole of any gas occupies 22.4 L
Therefore, 0.5 moles of the gas will occupy 22.4 x 0.5 L = 11.2 L

Therefore, the answer is – option (b).

Additional Information: Vapour density is the ratio of molar masses and is therefore unitless.

Note: Air has a vapour density of one. Therefore, we can also say that vapour density is an indication of whether any gas is more (density > 1) or less (density < 1) denser than air.