Question

The partial pressures of CH4, N2, and O2 in a sample of gas were found to be 143 mmHg, 469 mmHg, and 563 mmHg, respectively. Calculate the mole fraction of oxygen. 20.1 0.399 0.741 0.479 0.359

Answer 1

Answer: The mole fraction of oxygen gas is 0.479

Explanation:

We are given:

Partial pressure of methane = 143 mmHg

Partial pressure of nitrogen gas = 469 mmHg

Partial pressure of oxygen gas = 563 mmHg

Total pressure = (143 + 469 + 563) = 1175 mmHg

To calculate the mole fraction of oxygen gas, we use the equation given by Raoult's law, which is:

$p_{O_2}=p_T\times \chi_{O_2}$

where,

$p_{O_2}$ = partial pressure of oxygen gas = 563 mmHg

$p_T$ = total pressure = 1175 mmHg

$\chi_{O_2}$ = mole fraction of oxygen gas = ?

Putting values in above equation, we get:

$563mmHg=1175mmHg\times \chi_{O_2}\\\\\chi_{O_2}=\frac{563}{1175}=0.479$

Hence, the mole fraction of oxygen gas is 0.479