Agas mixture contains 1.25 g n2 and 0.85 g o2 in a 1.55 l c ontainer at 18 °c. calculate the mole fraction and partial pressure of each component in the gas mixture.

- mole fraction of nitrogen is 0.626 and mole fraction of oxygen is 0.374.

partial pressure of nitrogen is 0.687 atm and partial pressure of oxygen is 0.410 atm.

Solution:- Moles of nitrogen = 1.25 g x (1mol/28g) = 0.0446 mol

moles of oxygen = 0.85 g x (1mol/32g) = 0.0266 mol

Total moles of gases in the container = 0.0446 + 0.0266 = 0.0712

mole fraction of a gas = moles of gas/total moles of the gases

so, mole fraction of nitrogen = 0.0446/0.0712 = 0.626

mole fraction of oxygen = 0.0266/0.0712 = 0.374

Volume of the container is 1.55 L and the temperature is 18 degree C that is 18 + 273 = 291 K

From ideal gas law equation, PV = nRT

P = nRT/V

Partial pressure of nitrogen = (0.0446 x 0.0821 x 291)/1.55 = 0.687 atm

and the partial pressure of oxygen = (0.0266 x 0.0821 x 291)/1.55 = 0.410 atm

Moles are calculated as the ratio of given mass to the molar mass. The mole fraction and partial pressure of each component in the mixture of gases will be:

Nitrogen

Oxygen

Moles are the base units of a substance. It is calculated as the ratio of given mass to the molar mass.

Now, the moles fraction of nitrogen will be:

Similarly, moles of oxygen will be:

Now, Total moles of gases = 0.0446 + 0.0266 = 0.0712

Also,

Mole fraction of nitrogen:

Similarly, for oxygen:

Also, given that,

Substituting the values:

Partial pressure of nitrogen:

Partial Pressure of Oxygen:

Therefore, from the the ideal gas law equation and mole fraction, we can determine the number of moles and partial pressure of gases.

To know more about partial pressure and moles, refer to the following link:

link

these nuts

practical process described pest part of scientific investigation.

because practical knowledge is powerful then therotical knowledge.