Question

In an equilibrium mixture containing $N_2{O}_4\left(g\right)$ and $N{O}_2\left(g\right)$ at a certain temperature, the $N{O}_2$ is found to be $25\%$ by volume. The molecular weight of the equilibrium mixture is:

• A. $40.25$
• B. $80.5$
• C. $70.4$
• D. $46.0$

Answer 1

The correct option is B $80.5$

Option $B$ is correct.

In order to find the molecular weight of the equilibrium mixture

Formula related with it is: $\overline{MW}=Y_A{M}_A+Y_B{M}_B\to$ Mole fraction

Here 25% of $N{O}_2$ and 75% of $N_2{O}_4$

$\therefore$ $\overline{MW}$= $\left(0.25\frac{molN{O}_2}{mol}\right)\left(46g/molofN{O}_2\right)+\left(0.75\frac{mol{N}_2{O}_4}{mol}\right)\left(92g/molof{N}_2{O}_4\right)$

$\overline{MW}$= 0.25$\times$46 + 0.75$\times$92

$\overline{MW}$= 11.5 + 69= 80.5

Therefore the molecular weight of the equilibrium mixture is 80.5