Question #2beda

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Question #2beda

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Answer 1 · 2016-12-23T13:18:32

Relative atomic masses of

$N \to 14$ $N->14$

$O \to 16$ $O->16$

Relative molecular masses of

$N O_{2} \to 14 + 2 \cdot 16 = 46$ $NO_2->14+2*16=46$

$N_{2} O_{4} \to 2 \cdot 14 + 4 \cdot 16 = 92$ $N_2O_4->2*14+4*16=92$
Since $vapour density = \frac{1}{2} molecular mass$ $"vapour density" =1/2"molecular mass"$

Vapour densities of

$N O_{2} \to = \frac{46}{2} = 23$ $NO_2->=46/2=23$

$N_{2} O_{4} \to \frac{2 \cdot 14 + 4 \cdot 16}{2} = \frac{92}{2} = 46$ $N_2O_4->(2*14+4*16)/2=92/2=46$

Let x be the number of moles of $N O_{2}$ $NO_2$ in 100 moles mixture.

So the vapour density of the mixture will be

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

$\Rightarrow 23 x + 4600 - 46 x = 3830$ $=>23x+4600-46x=3830$

$\Rightarrow 23 x = 770$ $=>23x=770$

$\Rightarrow x = \frac{770}{23} \approx 33.48$ $=>x=770/23~~33.48$

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Question #2beda

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