Question

A 1.00 L flask is filled with 1.30 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.250 atm .

What is the partial pressure of argon, PAr, in the flask?

Answer 1

Before adding ethane:

Molar mass of Ar = 39.95 g/mol

mass of Ar = 1.30 g

we have below equation to be used:

number of mol of Ar,

n = mass of Ar/molar mass of Ar

=(1.3 g)/(39.95 g/mol)

= 3.254*10^-2 mol

we have:

V = 1.0 L

n = 0.0325 mol

T = 25.0 oC

= (25.0+273) K

= 298 K

we have below equation to be used:

P * V = n*R*T

P * 1 L = 0.0325 mol* 0.0821 atm.L/mol.K * 298 K

P = 0.7951 atm

Even if ethane is added, the pressure due to Ar will be same

So,

partial pressure of Ar = 0.7951 atm