The vapour density of `N_(2)O_(4)` at a certain temperature is `30`. Calculate the percentage dissociation of `N_(2)O_(4)` this temperature.

Options

A. 0.535
B. 0.6
C. 0.745
D. 0.645

Answer

A

Explanation

Normal molecular mass of `N_(2)O_(4) = 28 + 64 = 92`
Normal vapour density (V.D before dissociation)
`(D) =` Normal moleculer mass/`2 = 92/2 = 46`
Vapour density after dissociation `(d) = 30`
`N_(2)O_(4)(g) rarr 2NO_(2)(g)`
Degree of dissociation (x) = `D-d/d(n-1) = 46-30/30(2-1) = 16/30xx1 = 0.535 implies % = 53.5%`