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At equilibrium: N (2) O (4) hArr 2 NO (2...

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:

A

`10%`

B

`15%`

C

`20%`

D

`18%`

Text Solution

Verified by Experts

The correct Answer is:
B
Degree of dissociation may be calculated as,
`x = (M-m)/((n-1)m) because n = 2` (number of gas mole produced by 1 mol reactant)
`= (92-80)/((2-1)80) (M = 92, m = 80) = (12)/(80) = 0.15`
Percentage dissociation `= 0.15 xx 100 = 15`
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