A sample of oxygen gas was collected via water displacement. since the oxygen was collected via water displacement, the sample i

Question

3 years ago

7

A sample of oxygen gas was collected via water displacement. since the oxygen was collected via water displacement, the sample i

s 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.

Chemistry

1 answer:

mylen [45] · Answer 1 · 2022-02-11T12:47:28+03:00

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