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Answer 1 · 2016-12-23T13:18:32

Relative atomic masses of

$N->14$

$O->16$

Relative molecular masses of

$NO_2->14+2*16=46$

$N_2O_4->2*14+4*16=92$
Since $"vapour density" =1/2"molecular mass"$

Vapour densities of

$NO_2->=46/2=23$

$N_2O_4->(2*14+4*16)/2=92/2=46$

Let x be the number of moles of $NO_2$ in 100 moles mixture.

So the vapour density of the mixture will be

$(23*x+46*(100-x))/100=38.3$

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

$=>23x=770$

$=>x=770/23~~33.48$