The total pressure of a mixture of 604 g of `O_(2)` and 5.6 g of `N_(2)` present in a 2 L vessel is 1200 mm. What is the partial pressure (in mm) of nitrogen ?

Options

A. 1200
B. 600
C. 900
D. 200

Answer

b

Explanation

Number of moles of oxygen =`6.4/32` = 0.2
Number of moles of nitrogen = `5.6/28` = 0.2
`therefore` Mole fraction of `N_(2) = n/(n+N) =0.2/(0.2+0.2) = 1/2`
Partial pressure = Mole fraction of `N_(2) xx `Total pressure
`= 1/2 xx 1200 = 600` mm