Question

Nitrogen dioxide $\left( NO _{2}\right)$ cannot be obtained in a pure form in the gas phase at exists as a mixture of $NO _{2}$ and $N _{2} O _{4}$. At $298\, K$ and $0.98 \,atm$, the density of this gas mixture is $2.764 \,g\, L ^{-1}$. Thus partial pressure of $NO _{2}$ in the mixture is

• A. 0.40 atm
• B. 0.51 atm
• C. 0.49 atm
• D. 0.60 atm

Answer 1

Answer: (A)

$pV = nRT =\frac{w}{m} R T $

$\therefore p=\frac{w}{V} m R T $

$p=\frac{d}{m} R T$

$m$ (mixture ) $= \frac{d R T}{p}$

$=\frac{2.764 \times 0.0821 \times 298}{0.98} $

$=69 \,g \,mol ^{-1}$

Let mole fraction of $NO _{2}=x$ and that of $ N _{2} O _{4}=(1-x)$

Then $ \frac{M\left( NO _{2}\right) x+M\left( N _{2} O _{4}\right)(1-x)}{1+(1-x)}$

$46 x+92(1-x)=69.0=69.0\, g\, mol ^{-1} $

Thus, $ x=0.50 $

$p_{ NO _{2}}=0.50 \times p_{\text {total }}=0.49\, atm$