Question

At ${80}^{o}C$ and 2 atm, $N_2{O}_4$ is $75$% dissociated into $N{O}_2$. The approximate volume occupied by $18.4$ g of $N_2{O}_4$ under these conditions is:

• A. $5$ litres
• B. $6$ litres
• C. $7$ litres
• D. $4$ litres

Answer 1

Answer: (A)

$5$ litres

Under similar conditions of $T$ and total pressure, degree of dissociation $\left(\alpha \right)$ will remain the same.

$18.4g{N}_2{O}_4=\frac{18.4}{92}=0.2$

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

at $t$ $=$ $0,0.2$ mole

$t_{eq}=0.2(1-0.75)(2\times 0.2\times 0.75)$

Total number of moles $0.2(1+0.75)$

$=0.35$

$V=\frac{nRT}{P}$

$=\frac{0.35\times 0.0821\times 353}{2}$

$=5.07$

$\simeq 5.0$