Question

At $340$ K and $1$ atm pressure, $N_2{O}_4$ is $66\%$ dissociated into $N{O}_2$. What volume does $10$ g $N_2{O}_4$ occupy under these conditions?

• A. $5.023$ L
• B. $10.4$ L
• C. $15.4$ L
• D. $0.05$ L

Answer 1

The correct option is A $5.023$ L

The dissociation reaction is as follows:

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

Initial number of moles $1$ $0$

Number of moles at equilibrium $1-a$ $2a$

Here, $a$ is the degree of dissociation which is equal to $0.66$.

Total number of moles at equilibrium $=1-a+2a=1+a=1+0.66=1.66$

$1$ mole of $N_2{O}_4$ present initially corresponds to $1.66$ moles at equilibrium.

$\frac{10}{92}$ moles of $N_2{O}_4$ present initially corresponds to $\frac{1.66\times 10}{92}=0.18$ moles at equilibrium.

The ideal gas equation is $PV=nRT$.
Substituting values in the above expression, we get
$1\times V=0.18\times 0.0821\times 340$
$V=5.02$ L

Hence, the volume occupied by $10$ g of $N_2{O}_4$ will be $5.02$ L.