Answer:- 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
