Question

Part A

A gas mixture contains 1.35 g N2 and 0.89 g O2 in a 1.51-L container at 15 ∘C.

Calculate the mole fraction of N2 & O2

Calculate the partial pressure of N2 & O2

Part B

A 1.00 L flask is filled with 1.05 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.450 atm .

What is the partial pressure of argon, PAr, in the flask?

What is the partial pressure of ethane, Pethane, in the flask?

Answer 1

Part A)

Moles of N2 = n1 = mass/molar mass = 1.35/28 = 0.04821

Moles of O2 = n2 = mass/molar mass = 0.89/32 = 0.02781

Total moles = n = n1 + n2 = 0.04821 + 0.02781 = 0.076023

Mole fraction of N2 = x1 = n1/n = 0.04821/0.076023 = 0.6342

Mole fraction of O2 = x2 = n2/n = 0.02781/0.076023 = 0.3658

Let pressure of N2 is P1 and pressure of O2 is P2 atm.

Volume (V) = 1.51 liters

Gas constant (R) = 0.0821 L atm/mol K

Temperature (T) = 15°C = 15 + 273 = 288 K

Using ideal gas equation for N2;

P1 * V = n1 * R * T

P1 = n1 * R * T /V = 0.04821 * 0.0821 * 288 / 1.51 = 0.755 atm

Similarly;

P2 = n2 * R * T/V = 0.02781 * 0.0821 * 288 / 1.51 = 0.435 atm .....Answer

Part B)

Moles of Ar (n1) = mass/molar mass = 1.05/40 = 0.02625

Let pressure of Ar is P1 atm. Using ideal gas equation;

P1 = n1 * R * T / V = 0.02625 * 0.0821 * (25+273) /1 = 0.642 atm ....Answer

Let moles of ethane = n2

Pressure of ethane = P2 = 1.450 - 0.642 = 0.808

Using ideal gas equation for ethane;

P2 * V = n2 * R * T

n2 = P2 * V / R * T = 0.808 * 1 / 0.0821 * (25+273)

n2 = 0.033 moles

Mass of ethane = moles * molar mass = 0.033 * 30 = 0.99 grams = 1 gram (approx) ....Answer

Let me know if any doubts/if answer is not matching.