Question

At 340 K and 1 atm pressure,$N_2{O}_4$ is 66% dissociated into ${NO}_2$. What volume of 10g $N_2{O}_4$ occupy under these conditions?

• A. $V=10L$
• B. $V=2.5L$
• C. $V=5.04L$
• D. $V=1.5L$

Answer 1

Answer: (C)

$V=5.04L$

$N_2{O}_4\rightleftharpoons 2{NO}_2$
$10$ Moles before equilibrium
$(1-\alpha )$ $2\alpha$ Moles at equilibrium
(Let $\alpha$ be degree of dissociation)
$\therefore$ Total moles at equilibrium $=1+\alpha (\because a=0.66)$
$=1+0.66=1.66$
1 mol of $N_2{O}_4$ is taken, moles at equilibrium = 1.66
$\frac{10}{92}$ mol of $N_2{O}_4$ is taken, moles at equilibrium
$=\frac{1.66\times 10}{92}=0.18$
$\because PV=\frac{w}{m}RT$
$1\times V=0.18\times 0.0821\times 340$
$\therefore V=5.04L$