Question

A 1-L flask is filled with 1.45 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.05 atm .

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

b. What is the partial pressure of ethane, Pethane, in the flask?

Answer 1

a)

Molar mass of Ar = 39.95 g/mol

mass(Ar)= 1.45 g

use:

number of mol of Ar,

n = mass of Ar/molar mass of Ar

=(1.45 g)/(39.95 g/mol)

= 3.63*10^-2 mol

Given:

V = 1.0 L

n = 0.0363 mol

T = 25.0 oC

= (25.0+273) K

= 298 K

use:

P * V = n*R*T

P * 1 L = 0.0363 mol* 0.08206 atm.L/mol.K * 298 K

P = 0.8877 atm

Answer: 0.888 atm

b)

P ethane = P total - P argon

= 1.05 atm - 0.888 atm

= 0.162 atm

Answer: 0.162 atm