Answer:
1) The diometic gas is N2 (molar mass 28 g/mol)
2) The partialpressure of oxygen is 316.6 mmHg
Explanation:
Step 1: Data given
Volume = 4.8 L
pressure = 1.60 atm
temperature = 30.0°C
Difference in mass after weighting again = 8.7 grams
Step 2:
PV = nRT
⇒ with P = the pressure of the gas = 1.60 atm
⇒ with V = the volume of the gas = 4.8 L
⇒ with n = the number of moles = mass/molar mass
⇒ with R = the gas constant = 0.08206 L*atm/k*mol
⇒ with T = the temperature = 30.0 °C = 303 K
(1.60 atm) (4.8L) = (n)*(0.08206)*(303 K)
n = (1.60 * 4.8) / ( 0.08206*303)
n = 0.30888 mol
Step 3: Calculate molar mass
Molar mass = mass / moles
Molar mass = 8.7 grams / 0.3089 moles
Molar mass ≈ 28 g/mol
The diometic gas is N2
2) What is the partial pressure of oxygen in the mixture if the total pressure is 545mmHg ?
Step 1: Calculate mass of nitrogen
Let's assume a 100 gram sample. This means 38.8 grams is nitrogen
Step2: Calculate moles of N2
Moles N2 = mass N2 / molar mass N2
Moles N2 = 38.8 grams / 28 .02 grams
Moles N2 = 1.38 moles
Step 3: Calculate moles of O2
Moles O2 = (100 - 38.8)/ 32 g/mol
Moles O2 = 1.9125 moles O2
Step 4: Calculate molefraction of oxygen
Molefraction O2 = moles of component/total moles in mixture
=1.9125/(1.9125 + 1.38 moles)
=0.581
Step 5: Calculate the partial pressure of oxygen
PO2 =molefraction O2 * Ptotal
=0.581 * 545mmHg
=316.6 mmHg
The partialpressure of oxygen is 316.6 mmHg
