Question

$N_2{O}_4$ is 20% dissociated at ${27}^{o}C$ and 760 torr. The density of the equilibrium mixture is:

• A. 3.1 g/L
• B. 6.2 g/L
• C. 12.4 g/L
• D. 18.6 g/L

Answer 1

Answer: (A)

3.1 g/L

Density, $d=\frac{(P.M_{av})}{RT}$

$N_2{O}_4$ is $20\%$ dissociated

$\underset{\text{x}\text{(x-0.2x)}}{N_2{O}_4}\leftrightharpoons \underset{\text{0}\text{(0.4x)}}{2N{O}_2}$
Moles of mixture at eqm. $=0.4x+0.8x$
$M_{av}=\frac{0.4x\times M_{N{O}_2}+0.8x\times M_{N_2{O}_4}}{(0.4+0.8)x}=\frac{0.4x\times 46+0.8x\times 92}{1.2x}=76.66g/mole$

$760torr=1atm,T=300K,R=0.082atm.L/mol/K$
$d=\frac{(1\times 76.66)}{0.082\times 300}=3.1g/L$