Question #2beda

Dec 23, 2016

Relative atomic masses of

$N \to 14$

$O \to 16$

Relative molecular masses of

$N {O}_{2} \to 14 + 2 \cdot 16 = 46$

${N}_{2} {O}_{4} \to 2 \cdot 14 + 4 \cdot 16 = 92$
Since $\text{vapour density" =1/2"molecular mass}$

Vapour densities of

$N {O}_{2} \to = \frac{46}{2} = 23$

${N}_{2} {O}_{4} \to \frac{2 \cdot 14 + 4 \cdot 16}{2} = \frac{92}{2} = 46$

Let x be the number of moles of $N {O}_{2}$ in 100 moles mixture.

So the vapour density of the mixture will be

$\frac{23 \cdot x + 46 \cdot \left(100 - x\right)}{100} = 38.3$

$\implies 23 x + 4600 - 46 x = 3830$

$\implies 23 x = 770$

$\implies x = \frac{770}{23} \approx 33.48$