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

Answer:

7.79 moles

Explanation:

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

Moles of helium = $\frac{x}{4}$ moles

Moles of argon = $\frac{x}{40}$ moles

Total moles = $\frac{x}{4}+\frac{x}{40}=\frac{11x}{40}\ moles$

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:

$3.50\times 80.0=\frac{11x}{40}\times 0.0821\times 398$

x=31.16 g

Moles of helium = 31.16 / 4 = 7.79 moles