Question

A sample of oxygen gas was collected via water displacement. since the oxygen was collected via water displacement, the sample is saturated with water vapor. if the total pressure of the mixture at 26.4 °c is 749 torr, what is the partial pressure of oxygen? the vapor pressure of water at 26.4 °c is 25.81 mm hg.

Answer 1

The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.

Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr

Vapor pressure of pure water at $26.4^{0}C$=25.81mmHg

                               = $25.81 mmHg*\frac{1 Torr}{1 mmHg} =25.81 Torr$

According to Dalton's law of partial pressures,

Total pressure = Partial pressure of Oxygen gas + Partial pressure of water

  749 Torr = Partial pressure of Oxygen gas + 25.81 Torr

Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr =  723.19 Torr

Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr