Answer:
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
