A mixture of `NO_(2)` and `N_(2)O_(4)` has a vapor density of `38.3` at 300 K. What is the number of moles of `NO_(2)` in 100 g of themixture ?

Options

A. `0*043`
B. `4*4`
C. `3*4`
D. ` 0*437`

Answer

D

Explanation

Suppose `NO_(2) = x g . " Then " N_(2) O_(4) = ( 100 - x) g `
`" Moles of " NO_(2)= x/46,"Moles of " N_(2)O_(4)=(100-x)/92`
`"Mole fraction of "NO_(2)(x//46)/(x//46 + ( 100 - x ) //92)`
`= x/46 xx92/(100+x)=(2x)/(100+x)`
Mole fraction of `N_(2)O_(4) = 1 - (2x)/(100+x) = (100 - x)/(100+x) `
Molar mass of mixture
`= (2x)/(100 + x) xx 46 + ( 100-x)/(100+x) xx92 = 9200/(100+x) `
` :. 9200/(100+x) = 2 xx 38*3 = 76*6`
or` 76*6 x = 9200 = 1540 or x = 20*10 g `
` :." Moles of " NO_(2) = (20*10)/46 = 0. 437`