Question

Part A A mixture of He, Ar, and Xe has a total pressure of 2.20 atm...

Part A

A mixture of He, Ar, and Xe has a total pressure of 2.20 atm . The partial pressure of He is 0.450 atm , and the partial pressure of Ar is 0.400 atm . What is the partial pressure of Xe? Express your answer to three significant figures and include the appropriate units.

Part B

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? Express your answer to three significant figures and include the appropriate units.

Homework Answers

Answer #1

Part A:

Given mixuture of gases has total pressure,

Ptotal=2.20 atm=P(He) + P(Ar) + P(Xe)

Given P(He)=0.45 atm, P(Ar)=0.4 atm, and P(Xe)=?

According to Dalton's law of partial pressures,

P(total)=P(He) + P(Ar) + P(Xe)

2.20 atm=0.45 atm + 0.4 atm + P(Xe)

P(Xe)=2.2 atm - 0.85 atm

P(Xe)=1.35 atm.

Part B:

Given 0.25 mol N2, 0.25 mol O2 and unknown moles of He.

Temperature, T=0°C=273 K, pressure, P=1 atm, volume, V=18 L, gas constant R=0.0821 L atm mol^-1 K^-1.

From Ideal gas equation, PV=nRT

where n=number of moles

n=PV/RT

n=(1 atm x 18 L)/(0.0821 L atm mol^-1 K^-1 x 273 K)

n=0.803 mol.

Total number of moles =n(N2) + n(O2) + n(He)=0.803

n(He) =0.80 3 - 0.25 -0.25 mol=0.303 mol.

Molar mass of He=4g/mol.

Therefore mass of He=moles x molar mass=0.303 mol x 4 g/mol=1.21 g.

Please let me know if you have any doubt. Thanks.

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