Question

Vapour density of gas is $11.2$ , volume occupied by $2.4g$ of this at STP will be :
A. $11.2L$
B. $2.24L$
C. $22.4L$
D. $2.4L$

Answer 1

The vapour density of a gas is the density of a vapour compared to that of hydrogen. The STP stands for standard temperature and pressure which is $273K$ and $1atm$ .

Complete answer:
The vapour density is the ratio of the mass of n molecules of gas and molar mass of hydrogen. The mass of n molecules of gas is the molar mass of the gas and the molar mass of hydrogen is $2.016g$ . Thus the vapour density of a gas is expressed as
$vapour{\text{ }}density = \dfrac{{molar{\text{ }}mass}}{{2.016}}$
Or $vapour{\text{ }}density = \dfrac{1}{2}molar{\text{ }}mass$
Given the vapour density of a gas is $11.2$. Thus the molecular mass of the gas is calculated using the above equation.
$molar{\text{ }}mass{\text{ }}of{\text{ }}gas = 2 \times vapour{\text{ }}density$
$= 2 \times 11.2 = 22.4g$
The mass of gas given is $2.4g$ at STP. Thus the number of moles of gas present in $2.4g$ of gas is
$= \dfrac{{mass{\text{ }}of{\text{ }}gas}}{{molar{\text{ }}mass}}$
$= \dfrac{{2.4g}}{{22.4g}} = 0.107moles$.
At STP means at $273K$ temperature and $1atm$ pressure, the volume of gas occupied by one mole of gas is $22.4L$ . The mole of gas present here is $0.107moles$ .
Thus, the volume occupied in $0.107moles$ of gas = $22.4 \times 0.107 = 2.4L.$
Hence option D is the correct answer, i.e. volume occupied by $2.4g$ of this at STP will be $2.4L$.

Note:
The ideal gas law indicates that the volume occupied by a gas solely depends on the amount of gas and temperature and pressure of the gas. Standard temperature and pressure conditions are abbreviated as STP, which signifies the temperature is $0^\circ C$ and pressure is $1atmosphere$. Parameters of gases are very important during the calculation of other unknown variables related to the gas.