Vapour density of the equilibrium mixture of `NO_(2) and N_(2) O_(4)` is found to be 40 for the equilibrium: `N _(2) O _(4) (g) hArr 2 NO _(2) (g).` Calculate percentage of `NO_(2)` in the mixture.

Options

A. `26.08%`
B. `21.52%`
C. `19.24%`
D. `24.62%`

Answer

A

Explanation

`{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial moles",1,,0),("Moles at equilibrium",(1-x),,2x):}`
Hence, the total moles at equilibrium `= (1+x) = 1+0.15 = 1.15`
`("moles of " NO_(2))/("total mass") = (2 xx 0.15)/(1.15) xx 100`, Percentage of `NO_(2) = 26.08%`