Question

Experiment 3: Measure the Mass of Air

Took a 150 mL Erlenmeyer flask & a balance & placed them on the workbench.

Moved the flask onto the balance; the mass = 88.000 g

Closed the flask & attached a pressure gauge to the Erlenmeyer flask.

Weight of the closed Erlenmeyer flask = 88.179 g

Added 1.50 atm to the Erlenmeyer flask.

The pressure = 1.50 atm

The mass = 88.269 g

Added air to the Erlenmeyer flask to a pressure of 2.00 atm.

The pressure = 2.00 atm

The mass = 88.358 g

Added air to the Erlenmeyer flask to a pressure of 2.50 atm.

The pressure = 2.50 atm

The mass = 88.448 g

Added air to the Erlenmeyer flask to a pressure of 3.00 atm.

The pressure = 3.00 atm

The mass = 88.538   

How many moles of air were in the Erlenmeyer flask when the total air pressure was 2.00 atm?

A. 0.179 mol

B. 0.0124 mol

C. 0.00620 mol

D. 0.0155 mol

How many moles of air were in the Erlenmeyer flask when the total air pressure was 2.00 atm?

A. 0.179 mol

B. 0.0124 mol

C. 0.00620 mol

D. 0.0155 mol

Answer 1

Air is mainly nitrogen and oxygen. Its approximate composition is 78% N₂ and 22% O₂. Weight of one mole of N₂ is 28g and weight of one mole of O₂ is 32g. So the approimate weight of one mole of air will be (28 X 78 + 32 X 22)/100 = 28.88g

Weight of air in the Erlenmeyer flask when the pressure is 2.00 atm. = 88.358g - 88.179g = 0.179g

Number of moles of air present when the the air weight is 28.88g = 1

Number of moles of air present when the air weight is 0.179g = 0.179/28.88 = 0.00620

Hence the number of moles of air in the Erlenmeyer flask at 2.00 atm. total air pressure will be 0.00620

So the answer is (C).