Question

A 100 mL sample of oxygen gas is collected over water at 26 °C and at a pressure of 756 torr. Given that the vapor pressure of water at 26 °C is 25 torr, how many moles of dry oxygen are present?
Show all steps and explain fully.
ANSWER: 3.9 x 10^-3 mol

Answer 1

The volume of gas collected = 100mL

pressure of wet gas = 756 torr

vapor pressure of water = 25 torr

When a gas is collected over water , it is saturated with water vapors. The pressure due to water vapors at that temperature to be substracted from the measured pressure to get the pressure of dry gas.

However the volume of gas is not effected by the presence of saturated vapor pressure.

Thus pressure of dry gas = 756-25 = 731 torr

= 731/760 atm

temperature = 26C = 26+273 = 299K

Using the ideal gas equation PV =nRT

number of moles of O2 gas n = PV /RT

= { (731/760)atm x 0.1L} / 0.0821L.atm/K.mol x 299K

= 3.918x 10^-3 mol