Question

A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg?

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Answer 1

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{O_2}$ = partial pressure of $O_2$ = ?

$\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}$

$P_{T}$ = total pressure of mixture  = 525 mmHg

${\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles$

${\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles$

Total moles = 1.94 + 1.35 = 3.29 moles

$\chi_{O_2}=\frac{1.94}{3.29}=0.59$

$p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg$

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Answer 2

A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.

A gaseous mixture of O₂ and N₂ contains 37.8% nitrogen by mass, that is, in 100 g of the mixture, there are 37.8 g of N₂. The mass of O₂ in 100 g of the mixture is:

$mO_2 = 100 g - 37.8 g = 62.2 g$

We will convert both masses to moles using their molar masses.

$N_2: 37.8 g \times 1 mol/28.00 g = 1.35 mol\\\\O_2: 62.2 g \times 1 mol/32.00 g = 1.94 mol$

The mole fraction of O₂ is:

$\chi(O_2) = \frac{nO_2}{nN_2+nO_2} = \frac{1.94mol}{1.35mol+1.94mol} = 0.590$

Given the total pressure (P) is 525 mmHg, we can calculate the partial pressure of oxygen using the following expression.

$pO_2 = P \times \chi(O_2) = 525 mmHg \times 0.590 = 310 mmHg$

A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.

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