Answer:
0.178 moles of oxygen gas molecules were collected.
Explanation:
Pressure at which oxygen gas collected = p= 1.00 atm
Vapor pressure of the water [tex]p'=34.5 Torr=\frac{34.5}{760} atm=0.0454 atm[/tex]
Pressure of the oxygen gas = P
p = p' + P
[tex]1.00 atm =0.0454 atm+P[/tex]
P = 1.00 atm - 0.0454 atm = 0.9546 atm
Volume of the oxygen gas = V = 4.56 L
Moles of oxygen gas = n
Temperature at which gas collected = T = [tex]25^oC=273+25=298 K[/tex]
[tex]PV=nRT[/tex] (ideal gas)
[tex]n=\frac{PV}{RT}=\frac{0.9546 atm\times 4.56 L}{0.0821 atm L/mol K\times 298 K}=0.178 mol[/tex]
0.178 moles of oxygen gas molecules were collected.
