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?

Options

A. `2.8xx10^(-4)g`
B. `5.2xx10^(-3) g`
C. `1.4xx10^(-2) g`
D. `0.07 g`

Answer

C

Explanation

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`