Question

n these partial pressures: 3.5 atm n2, 2.8 atm o2, 0.25
$p_t = p_1 + p_2 + p_3 + ... + p_n$
$\frac{p_a}{p_t} = \frac{n_a}{n_t}$
what is the mole fraction of o₂ in the mixture?
□ atm
what is the mole fraction of ar in the mixture?
□ atm
done

Explanation:

Step1: Calculate total pressure

$P_T = P_{N_2} + P_{O_2} + P_{Ar} = 3.5\ \text{atm} + 2.8\ \text{atm} + 0.25\ \text{atm} = 6.55\ \text{atm}$

Step2: Find mole fraction of $\text{O}_2$

Use $\frac{P_a}{P_T} = \frac{n_a}{n_T}$, so $\chi_{O_2} = \frac{P_{O_2}}{P_T}$
$\chi_{O_2} = \frac{2.8}{6.55} \approx 0.4275$

Step3: Find mole fraction of $\text{Ar}$

Use $\frac{P_a}{P_T} = \frac{n_a}{n_T}$, so $\chi_{Ar} = \frac{P_{Ar}}{P_T}$
$\chi_{Ar} = \frac{0.25}{6.55} \approx 0.0382$

Answer:

Mole fraction of $\text{O}_2$: $\approx 0.43$
Mole fraction of $\text{Ar}$: $\approx 0.038$