Answer to Question #263518 in General Chemistry for diamond

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

26.4 °C is 741

741 torr, what is the partial pressure of oxygen? The vapor pressure of water at 26.4

26.4 °C is 25.81

25.81 mm Hg.

Answer 1

The total pressure exerted by a wet gas is equal to the sum of the partial pressure of the gas itself + the vapor pressure of water at that temperature.

$\implies p_{total}=p_{H_20}+p_{gas}$

Given $p_{gas}=25.81mm Hg$ ;$p_{total}=741torr=741mmHg$

$\implies p_{gas}=741-25.81=715.19mmHg$