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The vapour density of a mixture of NO_(2) and N_(2)O_(4) is 38.3 at 26.7^(@)C. Calculate the number of moles of NO_(2) in 100 g of the mixture.

Answer»

Solution :Suppose `NO_(2)` present in 100 g of the mixture = x g
Then `N_(2)O_(4)` present in the mixture `=(100-x)g`
MOLAR MASS of `NO_(2)=14+32="46 g mol"^(-1)`
`"Molar mass of "N_(2)O_(4)="92 g mol"^(-1)`
`"Molar mass of mixture "=2xxV.D.=2xx38.3="76.6 g mol"^(-1)`
Expressing all quantities in TERMS of moles, we should have
`(x)/(46)+(100-x)/(92)=(100)/(76.6)"or"92x+4600-46x=5524.8`
`"or"46x=924.8"or"x=20.1g`
` therefore"No. of moles of NO"_(2)" in the mixture"=(20.1)/(46)=0.437`


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