Why is air transparent? I made a rough calculation (probably having made a miscalculation though) following here and here to find out the number of air molecules in 1 cubic centimetre of air. 
The first website says there are about $0.02504\times10^{27}$ air molecules in 1 cubic metre. The the second one says the diameter of an air molecule is about $4\times 10^{-10}$m. Then the area of the molecule is about $4\pi\times 10^{-20} \approx 12\times 10^{-20}$.
Then we get the number of air molecules in one cubic metre to be about $12\times0.025\times10^7$, and in one cubic centimetre to be about $12\times0.025\times10^5 = 30000$.
So if there are 30000 air molecules in 1 cubic centimetre, it's almost impossible for light to miss hitting all of them. Following this logic, I can't comprehend how light propagates in air (I would expect it to be scattered, absorbed etc). How does physics explain this? 
Furthermore, since the speed of light in a vacuum is constant, but light is "slower" in higher refractive index materials, can I assume light travels a longer path when it travels through a medium with a refractive index higher than empty space? If not, what decreases its speed? 
(I thought two questions are related as they both question how light travel in air.)
 A: Firstly, the density of air is pretty low - the light is unlikely to collide with a particle. However, this is not the most important reason. 
All elements have a spectra, and it happens to be that those which make up air do not absorb frequencies which are visible to the human eye. Therefore, the light they re-emit is not visible to us, thus making air seem to be transparent.
A: It may be noted that although the classical cross section come out to be too high but it does not mean that the light will be fully absorbed because the light interacts with an atom via Rayleigh scattering process which is rather weak.  
Your argument may be valid if each atom fully absorb light but every atom is almost transparent to the light (only interact via weak Rayleigh scattering) hence light is transmitted. 
On the other hand if you use a radiation which is having wavelength in resonance with the gas transitions in air they may get efficiently absorbed or if you reduce the wavelength below 200 nm the radiation become ionizing (photon energy > ionization potential of gases) then also it will get absorbed very efficiently. 
