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_(H_(2)O) + p_(CO_(2)) + p_(N_(2)) = 1.74 mm`
`p_(CO_(2)) + p_(N_(2)) = 1.32 mm`
`p_(N_(2)) = 0.53 mm`
`therefore p_(CO_(2)) = 1.32 - 0.53 - 0.79 mm`
and `p_(H_(2)O) = 1.74 - 1.32 = 0.42 mm`.
Number of moles of each constituent is calculated using the equation
`pV = nRT`.
For `H_(2)O : pH_(H_(2)O) = (0.42)/(760)` atm, `V = (0.731)/(1000)` lit.
`T = 27 + 273 = 300 K, R = 0.082` lit. atm/K/mol
`(0.42)/(760) xx (0.731)/(1000) = n_(H_(2)O) xx 0.082 xx 300`
`n_(H_(2)O) = 1.64 xx 10^(-8)`
Similarly,
for `CO_(2) : (0.79)/(760) xx (0.731)/(1000) = n_(CO_(2)) xx 0.082 xx 300`
`n_(CO_(2)) = 3.08 xx 10^(-8)`
and for `N_(2) : (0.53)/(760) xx (0.731)/(1000) = n_(N_(2)) xx 0.082 xx 300`
`n_(N_(2)) = 2.07 xx 10^(-8)`