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. 106.6%
• C. 26.7%
• D. 83.4%

Answer 1

The correct option is A 53.3%

$N_2{O}_4..............N{O}_2$, initially $N_2{O}_4$ has conc $=1$ than at equilibrium $N_2{O}_4$ has conc. $=2x$ initially we have taken $N_2{O}_4=1$ Molar mass $=92.V.D$ at eq $=30.M.V=60$
$n$ initial / $n$ final $=M$ final / $M$ initial
$1/1-X+2X=60/92$
$1/1+X=60/92$
$1+X=92/60$
$X=92/60-1$
$X=32/60$
$X=0.53$
percentage dissociation $=0.53\times 100=53$