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 mm Hg at 27^(@)C. The sample was transferred to a bulb in contact with dry ice (-75^(@)C) so that 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 (-195^(@)C) to freeze out the CO_(2). In the measured volume, the pressure was 0.53 mm Hg. How many moles of each constituents are in the mixture ?

A mixture of H_(2)O vapour, CO_(2) and N_(2) was trapped in a glass ap

1.	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 mm Hg at 27^(@)C. The sample was transferred to a bulb in contact with dry ice (-75^(@)C) so that 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 (-195^(@)C) to freeze out the CO_(2). In the measured volume, the pressure was 0.53 mm Hg. How many moles of each constituents are in the mixture ?
Answer» Solution :`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)`