Question

A mixture of helium gas and argon gas occupies 80.0 L at 398 K and 3.50 atm. If the mass of helium gas is equal to the mass of argon gas in the mixture, how many moles of helium does the mixture contain

Answer 1

7.79 moles

Step-by-step explanation:

Let the mass of helium gas = Mass of argon gas = x g

Moles of helium =
`(x)/(4)` moles

Moles of argon =
`(x)/(40)` moles

Total moles =
`(x)/(4)+(x)/(40)=(11x)/(40)\ moles`

Given that:

Temperature = 398 K

V = 80.0 L

Pressure = 3.50 atm

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

`3.50* 80.0=(11x)/(40)* 0.0821* 398`

x=31.16 g

Moles of helium = 31.16 / 4 = 7.79 moles