A gaseous mixture of O2 and N2 contains 34.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 465 mmHg?
This is incorrect, It says "by mass" not "by moles". therefore you must find the number of moles and use the mole fraction
First you find the moles of N2 assuming that you have 34.8 grams of N2. Then you find the moles of O2 assuming that you have 65.2 grams of O2.(100%-34.8%=65.2% so there is 65.2g O2). The partial pressure equals mole fraction x the total pressure(mole fraction= moles of O2/total moles in mixture) so in the end you multiply 465mmHg x mole fraction and theres your answer. :)
To find the partial pressure of oxygen in the mixture, we first need to determine the partial pressure of nitrogen.
Given that the gaseous mixture contains 34.8% nitrogen by mass, we can assume that the remaining 65.2% is oxygen. This means that the molar ratio of nitrogen to oxygen in the mixture is 34.8:65.2 or approximately 1:1.87.
Next, we can use Dalton's Law of Partial Pressures, which states that the total pressure exerted by a gaseous mixture is the sum of the partial pressures of each component gas. In this case, the total pressure is given as 465 mmHg.
Let's assume that the partial pressure of nitrogen is P_N and the partial pressure of oxygen is P_O2.
According to the molar ratio we determined earlier, the partial pressure of nitrogen can be calculated as follows:
P_N = (34.8/100) * 465 mmHg
P_N = 161.82 mmHg
Since the total pressure is the sum of the partial pressures of nitrogen and oxygen, we can write the equation as:
P_N + P_O2 = 465 mmHg
Substituting the value of P_N, we can solve for P_O2:
161.82 mmHg + P_O2 = 465 mmHg
P_O2 = 465 mmHg - 161.82 mmHg
P_O2 = 303.18 mmHg
Therefore, the partial pressure of oxygen in the mixture is approximately 303.18 mmHg.
0.348 x 465 mm = partial pressure N2.
partial pressure O2 = 465 - partial pressure N2 OR
partial pressure oxygen = 465 x (1.00 - 0.348) =