A gaseous mixture of and contains 30.8 nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 505mmHG ?
I am getting .41 * (505/750)= .27
.27 is my answer, is this right.
30.8 percent N2 by mass
69.2 precent O2 by mass
Now you need to convert those to moles
Assume 1000 g total
308/28=moles of N2
692/32 moles of O2
Now do those moles as a percent of total moles.
Then, multiply 505mmHg by the percent of O2 moles.
To find the partial pressure of oxygen in the gaseous mixture, you need to consider the nitrogen's mass percentage and its contribution to the total pressure.
First, calculate the mass percentage of nitrogen in the mixture.
Mass percentage of nitrogen = (mass of nitrogen / total mass of mixture) × 100
Let's assume the mass of nitrogen is "m" grams. The mass of the oxygen would then be (100 - m) grams.
Given that the mass percentage of nitrogen is 30.8%, we can write the equation:
30.8 = (m / (m + (100 - m))) × 100
Simplifying this equation:
30.8 = (m / 100) × 100
30.8 = m
So, the mass of nitrogen is 30.8 grams, and the mass of oxygen is (100 - 30.8) = 69.2 grams.
Next, calculate the partial pressure of oxygen.
Partial pressure of oxygen = (mass of oxygen / total mass of mixture) × total pressure
Using the given total pressure of 505 mmHg, the equation becomes:
Partial pressure of oxygen = (69.2 / (30.8 + 69.2)) × 505
Partial pressure of oxygen = (69.2 / 100) × 505
Partial pressure of oxygen = 0.692 × 505
Partial pressure of oxygen = 349.86 mmHg
Therefore, the partial pressure of oxygen in the mixture is approximately 349.86 mmHg.