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Answer :
Number of moles of oxygen = 0.037 mol
Given data:
Total pressure = 98.5 KPa
Partial pressure of nitrogen = 22.0 KPa
Partial pressure of argon = 50.0 KPa
Volume = 3.5 L
Temperature = 25°C (25+273= 298K)
Number of moles of oxygen = ?
Total pressure = P(N₂) + P(O₂) + P(Ar)
98.5 KPa = 22.0 KPa +P(O₂) + 50.0 KPa
98.5 KPa = 72.0 KPa +P(O₂)
P(O₂) = 98.5 KPa - 72.0 KPa
P(O₂) = 26.5 KPa
KPa to atm:
26.5 KPa/ 101 = 0.262 atm
Number of moles of oxygen:
PV = nRT
n = PV/RT
n = 0.262 atm × 3.5 L / 0.0821 atm.L/mol.K × 298 K
n = 0.917atm.L /24.47atm.L/ mol
n = 0.037 mol
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Answer:
Number of moles of oxygen = 0.037 mol
Explanation:
Given data:
Total pressure = 98.5 KPa
Partial pressure of nitrogen = 22.0 KPa
Partial pressure of argon = 50.0 KPa
Volume = 3.5 L
Temperature = 25°C (25+273= 298K)
Number of moles of oxygen = ?
Solution:
Total pressure = P(N₂) + P(O₂) + P(Ar)
98.5 KPa = 22.0 KPa +P(O₂) + 50.0 KPa
98.5 KPa = 72.0 KPa +P(O₂)
P(O₂) = 98.5 KPa - 72.0 KPa
P(O₂) = 26.5 KPa
KPa to atm:
26.5 KPa/ 101 = 0.262 atm
Number of moles of oxygen:
PV = nRT
n = PV/RT
n = 0.262 atm × 3.5 L / 0.0821 atm.L/mol.K × 298 K
n = 0.917atm.L /24.47atm.L/ mol
n = 0.037 mol
