Question

1. To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.8-L bulb, then filled it with the gas at 1.60atm and 30.0 ?C and weighed it again. The difference in mass was 8.7g . Identify the gas.2. A gaseous mixture of O2 and N2 contains 38.8{\rm \\%} nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 545mmHg ?

Answer 1

1) The diometic gas is N2 (molar mass 28 g/mol)

2) The partialpressure of oxygen is 316.6 mmHg

Step-by-step 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