Question

A 1.00 L flask is filled with 1.10 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.200 atm .

1. Part A

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

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

2. Part B

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

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

Answer 1

Part A

P_Ar = 0.674 atm

Explanation

Number of moles of argon= 1.10g/ 39.948g/mol = 0.027536mol

Total number of moles is calculated by ideal gas equation

PV = nRT

P = Pressure, 1.200atm

V = Volume , 1.00L

n = number of moles

R = gas constant , 0.082057 (L atm/mol K)

T = Tempeeature , 25℃ = 298.15K

n = PV/RT

= 1.200 atm × 1.00L / (0.082057 (L atm/mol K) × 298.15K)

= 0.049049mol

moles fraction = number of moles/total number of moles

mole fraction of argon = 0.027536mol/0.049049mol = 0.5614

Partial pressure = mole fraction × total pressure

Partial pressure of argon = 0.5614 × 1.200atm = 0.674atm

Part B

0.526 atm

Explanation

Mole fraction of ethane = 1 - mole fraction of argon

mole fraction of ethane = 1 - 0.5614 = 0.4386

Partial pressure of ethane = 0.4386 × 1.200atm = 0.526atm