A mixture of `H_(2)O` vapour, `CO_(2)` and `N_(2)` was trapped in a glass apparatus with a volume of `0.731 mL`. The pressure of the total mixture was `1.74 atm` at `27^(@)C`. The sample was transferred to a bulb in contact with dry ice `(-75^(@)C)` so that the `H_(2)O` vapour was frozen out. When the sample was returned to the measured volume, the pressure was `1.32 mm Hg`. The sample was then transferred to a bulb in contact with liquid nitrogen `(-95^(@)C)` to freeze out the `CO_(2)`. On the measured volume, the pressure was `0.53 mm Hg`. How many moles of each consituent there are in the mixture?

Explanation

`P_(N_(2))+P_(O_(2))+P_(H_(2)O)=1.74 atm` (`i`)
(`a`) When in contact with `-75^(@)C`, `H_(2)O` is frozen, so the pressure is due to `N_(2)` and `CO_(2)`
`P_(CO_(2))+P_(N_(2))=1.32 atm` (`ii`)
(`b`) When is contact with `-95^(@)C CO_(2)` is frozen out, so the pressure is due to only `N_(2)`
`P_(N_(2))=0.53 atm` (`iii`)
From equation (`i`), (`ii`) and (`iii`), we get
`P_(N_(2))=0.53 atm`, `P_(CO_(2))=0.79 atm`, `P_(H_(2)O)=0.42 atm`
Now, find moles using
`n=[(PV)/(RT)]`
`implies n_(N_(2))=1.57xx10^(-5)`, `n_(H_(2)O)=1.246xx10^(-5)`
and `n_(CO_(2))=2.34xx10^(-5)`