A volume of 18.0 l contains a mixture of 0.250 mole n2, 0.250 mole o2, and an unknown quantity of he. the temperature of the mixture is 0 ∘c, and the total pressure is 1.00 atm. how many grams of helium are present in the gas mixture?

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