At equilibrium: `N _(2) O _(4) hArr 2 NO _(2 (g))` the observed molecular weight of `N _(2) O _(4)` is `80 g mol ^(-1)` at 350 K. The percentage dissociation of `N _(2) O _(4 (g))` at 350 K is:

Options

A. `10%`
B. `15 %`
C. `20%`
D. `18%`

Answer

B

Explanation

Degree of dissociation may be calculated as
`x = (m - m )/( (n -1) m) because n =2` (number of gas moles produced by 1 ml reactant)
`= ( 92-80)/((2-1) 80) = (M = 92, m = 80) =(12)/(80) = 0.15`
Percentage dissociation `= 0.15 xx 100 = 15`