Question

The vapour density of $N_2{O}_4$ at a certain temperature is 30. What is the percentage dissociation of $N_2{O}_4$ at this temperature?

• A. 53.3 %
• B. 76.6 %
• C. 26.7 %
• D. None of the above

Answer 1

The correct option is A 53.3 %

The relationship between molar mass and vapour density is Molar mass $=2\times$ vapour density $=2\times 30=60$

	$N_2{O}_4$	$N{O}_2$
Initial moles	1	0
Moles at equilibrium	$1-\alpha$	$2\alpha$
Mole fraction	$\left(\frac{1-\alpha }{1+\alpha }\right)$	$\left(\frac{2\alpha }{1+\alpha }\right)$

Thus $92\left(\frac{1-\alpha }{1+\alpha }\right)+46\left(\frac{2\alpha }{1+\alpha }\right)=60$

$\alpha =0.533$

The percentage dissociation of $N_2{O}_4$ is $53.3\%$.