Answer:- 1.22 g of helium are present in gas mixture.
Solution:- Given, total pressure = 1.00 atm
T = 0 + 273 = 273 K
V = 18.0 L
From ideal gas law equation, PV = nRT
P = nRT/V
Partial pressure of nitrogen = (0.250 x 0.0821 x 273)/18.0
partial pressure of nitrogen = 0.311 atm
similarly, partial pressure of oxygen = (0.250 x 0.0821 x 273)/18.0
partial pressure of oxygen = 0.311 atm
From Dalton's law of partial pressure, total pressure is the sum of partial pressures of the gases present in the container.
So, total pressure = (partial pressure of nitrogen + partial pressure of oxygen + partial pressure of helium)
partial pressure of helium = 1.00 atm - (0.311 atm + 0.311 atm)
partial pressure of helium = 1.00 atm - 0.622 atm
partial pressure of helium = 0.378 atm
n = pV/RT
So, moles of helium = (0.378 x 18.0)/(0.0821 x 273) = 0.304 mol
Now we can calculate the grams of helium on multiplying it's moles by it's molar mass.
0.304 mol x (4.00g/1mol) = 1.22 g
