Question

A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure of oxygen in this mixture?

a. 140.0 mm hg
b. 320.0 mm hg
c. 373.0 mm hg
d. 480.0 mm hg

Answer 1

Hello!

We have the following statement data:

Data:

`P_(Total) = 800 mmHg`

`P\% N_(2) = 60\%`

`P\% O_(2) = 40\%`

`P_(partial) = ? (mmHg)`

As the percentage is the mole fraction multiplied by 100:


`P = X_{ O_(2) }*100`

The mole fraction will be the percentage divided by 100, thus:
What is the partial pressure of oxygen in this mixture?


`X_{ O_(2) } = (P)/(100)`

`X_{ O_(2)} = (40)/(100)`

`\boxed{X_{ O_(2)} = 0.4}`


To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:

In relation to
`O_(2)` :


`(P O_(2) )/(P_(total)) = X_O_(2)`

`(P O_(2) )/(800) = 0.4`

`P_O_(2) = 0.4*800`

`\boxed{\boxed{P_O_(2) = 320\:mmHg}}\end{array}}\qquad\quad\checkmark`

Answer:
b. 320.0 mm hg