A 1.00 L flask is filled with 1.20 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.300 atm . What is the partial pressure of argon, PAr, in the flask?

Partial pressure of Ar = 0.73 atm

Explanation:

Given data:

Volume of flask = 1.00 L

Mass of Ar = 1.20 g

Temperature = 25 °C (25+273.15 K= 298.15 k)

Total pressure = 1.300 atm

Partial pressure of Ar = ?

Solution:

First of all we will calculate the number of moles of Ar:

Number of moles = mass/ molar mass

Number of moles = 1.20 g/ 39.95 g/mol

Number of moles = 0.03 mol

Pressure of Ar:

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

P×1.00 L =  0.03 mol ×  0.0821 atm.L /mol.K× 298.15 k

P = 0.73 atm.L /1.00 L

P = 0.73 atm

Total pressure = Partial Pressure of Ar + Partial pressure of ethane

1.300 atm = 0.73 atm + P(ethane)

P(ethane) = 1.300 atm - 0.73 atm

P(ethane) = 0.57 atm

wher is the rest of the question