1.

An equimolar mixture of Nitrogen gas and water vapours is taken in a 2 litre flask at 27^(@)C and 1.23 xx 10^(-2) atm. pressure. What is the mass of the gas at -27^(@)C?

Answer»

`2.8xx10^(-4)g`
`5.2xx10^(-3) g`
`1.4xx10^(-2) g`
`0.07 g`

Solution :The number of moles of gaseous mixture, `n=(PV)/(RT)=(1.23xx10^(-2) X^2)/(0.082xx300)`
`=1 xx10^(-3)` moles
As the mixture is equimolar.
`THEREFORE .^nN_2=0.5xx10^(-3)`
and `.^nN_2=0.5xx10^(-3)` moles
At `-27^@C` water vapours CHANGES to water solid
`therefore` mass of gas shall be only due to `N_2` gas `=28xx5xx10^(-4)=1.4xx10^(-2) g`


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