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Question: The density of air is \(0.00130g{{L}^{-1}}\). The vapour density of air will be: (A) \(0.00065\) ...

The density of air is 0.00130gL−10.00130g{{L}^{-1}}. The vapour density of air will be:
(A) 0.000650.00065
(B) 0.650.65
(C) 14.481614.4816
(D) 14.5614.56

Explanation

Solution

The density of a substance is the ratio of its mass and the volume occupied by the substance. The vapour density of a substance can be calculated by using the density at standard temperature and pressure. At first, the molar mass of the substance would be calculated by the value of density, and then the vapour density will be half the value of molar mass.

Complete step-by-step solution: In order to solve this question, we will first write the definition density, which is the measure of a substance in terms of mass per volume. We know that the value of density is already provided to us in the question which is, 0.00130gL−10.00130g{{L}^{-1}}.
Now, we will calculate the mass of 22.4L22.4L of air, because one mole of air at STP, meaning standard temperature pressure, occupies 22.4L22.4L amount of volume. So, we use the formula,
D=mVD=\dfrac{m}{V}
Where mm and VV symbolizes mass and volume of the air respectively and the DD represents the density of the air. Since the volume and density are known to us we will put both of them on one side of the equation and after rearranging the equation we get,
m=V×Dm=V\times D
So, now we will put the value of volume 22.4L22.4L and the density 0.00130gL−10.00130g{{L}^{-1}}, we get,
m=22.4L×0.00130gL−1m=22.4L\times 0.00130g{{L}^{-1}}
On solving the equation, we get,
m=29.12g/molm=29.12g/mol
We know that the molar mass of a substance is twice its vapour density. This equation can be mathematically expressed as,
m=2×VDm=2\times {{V}_{D}}
Where VD{{V}_{D}} is the vapour density of air.
Now substituting the value of molar mass, in the above equation, we get,
29.12g/mol=2×VD29.12g/mol=2\times {{V}_{D}}
Now, after solving the above equation, we get,
VD=14.56{{V}_{D}}=14.56
So, the value of vapour density came out to be 14.5614.56, which matches the last option.

Hence, the most appropriate option would be option (D).

Note: One mole of air at the conditions of standard temperature and pressure, occupies exactly 22.4L22.4L volume. The molar mass of a substance is equal to twice the value of vapour density of that substance.