Question
Question: The density of air is \(0.001293\,{\text{g/}}\,{\text{mL}}\). Its Vapour density is: A. \({\text{1...
The density of air is 0.001293g/mL. Its Vapour density is:
A. 143
B. 14.48
C. 1.43
D. 0.143
Solution
Vapour density is determined by dividing the molecular mass of vapour by two. Molecular mass will be determined by using the density formula. The volume can be taken as the standard volume occupied by any gas at STP.
Formula used: Vapourdensity = 2molecularmass
Complete step by step answer:
The density of vapour with respect to the hydrogen is defined as the Vapour density. Vapour density is unit less. At standard temperature and pressure, as all the gas have the same temperature and pressure so, the density becomes the function of molar mass only. The Vapour density and molecular mass is directly related at STP. The molecular mass is high means the gas has a high density or vice versa.
The formula to determine the Vapour density is as follows:
Vapourdensity = 2molecularmass
At STP one mole of a gas occupies 22.4L volume.
Convert the volume from L to mL as follows:
1L=1000mL
⇒22.4L=22400mL
Determine the mass of the vapour as follows:
density = volumemass
Substitute 22400mL/molfor volume and 0.001293g/mL for density.
⇒0.001293g/mL = 22400mL/molmass
⇒mass=0.001293g/mL×22400mL/mol
⇒mass=28.96g/mol
So, the mass of vapour is 28.96g/mol
Determine the Vapour density is as follows:
Substitute 28.96 for molecular mass
⇒Vapourdensity = 228.96
⇒Vapourdensity = 14.48
So, the Vapour density is 14.48.
**Therefore, option (B) 14.48 is correct.
Note: **
STP is known as standard temperature and pressure. The value of standard temperature is 273K. The value of standard pressure is 1atm. One mole of a gas at 273K temperature and 1atm pressure occupies 22.4L volume. The Vapour density is used to determine whether a gas is more or less dense than the air.