Question

A sample of oxygen was collected in a tube over water. In the tube, there is a mixture of oxygen gas and gaseous water (steam). The total pressure of the gas mixture is 760. Torr. If the pressure of gaseous water is 23. 8 torr, what is the pressure of the pure oxygen? 99. 8 torr 736 torr 740. Torr 783 torr.

Answer 1

Considering the Dalton's law, the pressure of the pure oxygen is 736.2 torr.

Dalton's law

The pressure exerted by a particular gas in a mixture is known as partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

$P_{T} =P_{1} +P_{2} +...+P_{n}$

where n is the amount of gases present in the gas mixture.

This relationship is due to the assumption that there are no attractive forces between the gases.

Pressure of the pure oxygen

In this case, you  know that there is a mixture of oxygen gas and gaseous water (steam). Then the total pressure of a gas mixture is

$P_{T} =P_{O_{2} } +P_{water}$

You know:

• the total pressure of the gas mixture is $P_{T} =$760 Torr.
• the pressure of gaseous water is $P_{water} =$23. 8 torr.

Replacing:

$760 torr=P_{O_{2} } +23.8 torr$

Solving:

$P_{O_{2} } =760 torr - 23.8 torr$

$P_{O_{2} } =$736.2 torr

Finally, the pressure of the pure oxygen is 736.2 torr.

Learn more about Dalton's law:

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Answer 2

Answer:

740 on edge

Explanation: