In a flask of volume V litres, 0.2 mol of oxygen 0.4 mol of nitrogen, 0.1 mole of `NH_(3)` and 0.3 mol of He are enclosed at `27^(@)C`. If the total pressure exerted by these non reacting gases is one atmosphere, the partial pressure exerted by nitrogen is

Options

A. 1 atm
B. 0.1 atm
C. 0.2 atm
D. 0.4 atm

Answer

D

Explanation

`x_(N_(2)) = (n_(N_(2)))/(n_(O_(2)) + n_(N_(2)) + n_(NH_(3)) + n_(He))`
`= (0.4)/(0.2 + 0.4 + 0.1 + 0.3) = (0.4)/(1.0) = 0.4`
Partial pressure of nitrogen, `p_(N_(2)) = x_(N_(2)) xx P`
Where P is the total pressure
`p_(N_(2)) = 0.4 xx 1 atm = 0.4 atm`