Question

A volume of 17.5 L contains a mixture of 0.272 mole N₂ , 0.251 mole O₂ , and an unknown quantity of He. The temperature of the mixture is 0 °C , and the total pressure is 1.007 atm . How many grams of helium are present in the gas mixture? Round your answer to the hundredths of a gram without units.

Answer 1

Assumig that the gases in a gaseous mixture behave ideally,an Ideal gas law states that,

PV = nRT

We have,

P = 1.007 atm,

V = 17.5 L

R = 0.0821 L.atm.K.mol

T = 0 ^oC = 273.15 K

n = # of moles of gases.

Let us calculate n using all known values in Ideal Gas equation,

1.007 aatm * 17.5 L = n * 0.0821 L.atm/K.mol * 273.15 K

n = 1.007 aatm * 17.5 L / 0.0821 L.atm/K.mol * 273.15 K

n = 0.786 mol

Now,

n = # of mole of N2 + # of mole of O2 + # of mole of He = 0.786 mol

0.272 mol + 0.251 mol + # of mole of He = 0.786

# of mole of He = 0.786 - 0.272 - 0.251

# of mole of He = 0.263 mol

Atomic mass of He = 4.00 g/mol

Mass of 0.263 mol He = 0.263 mol * 4.00 g/mol = 1.05 g

Mass f helium gas is 1.05 g.

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