At `0^@ C and 1.0 atm ( = 1.01 xx 10^5 N//m^2)` pressure the densities of air, oxygen and nitrogen are `1.284 kg//m^3, 1.429 kg//m^3 and 1.251 kg //m^3` respectively. Calculate the percentage of nitrogen in the air from these data, assuming only these two gases to be present.

Answer

A, B

Explanation

Let mass of nitrogen `= (m) g`. Then, mass of oxygen `= (100 - m) g` Number of moles of nitrogen, `n_1 = (m)/(28)` and number of moles of oxygen `n_2 = ((100 - m)/(32))`
For air
`rho = (p M)/(R T) = (p)/(R T)((m_1 +m_2)/(n_1 + n_2))`
`:. 1.284 = (1.01 xx 10^5)/(8.31 xx 273)[(100 xx 10^-3)/((m//28) +(100 -m)//32)]`
Solving this equation, we get
`m = 76.5 g`
This is also percentage of `N_(2)` by mass on air as total mass have taken is `100 g`.