Question

$N_2{O}_4\left(g\right)\to 2N{O}_2\left(g\right)$ at a constant temperature, vapour density of $N_2{O}_4$ is found to be 30. Calculate % dissociation of $N_2{O}_4$.

• A. 53.3%
• B. 46.7%
• C. 16.7%
• D. 83.3%

Answer 1

The correct option is A 53.3%

The given vapour density (d) is the observed one.

For degree of dissociation;

$N_2{O}_4\leftrightharpoons 2N{O}_2$

$\%\alpha =\frac{D-d}{d(y-1)}\times 100$

where,

D is calculated vapour phase = $2\times molarmass=92$

y is number of mole of product per mole reactant.

$\%\alpha =\frac{92-30}{30(2-1)}\times 100$

$\%\alpha =53.3\%$

Option A is correct answer