Gas Laws and Molar Mass Calculations

This video tutorial covers mixed mole conversions, reviewing key conversion factors such as 1 mole equals 6.02 * 10^23 particles, the molar mass in grams, and 22.4 liters. It includes sample problems on calculating the mass of nitrogen gas and the number of molecules in a carbon dioxide sample. Students are encouraged to solve practice problems to reinforce their understanding.

What is the value of Avogadro's number?

• 3.14 * 10^23

• 1.23 * 10^23

• 9.81 * 10^23

• 6.02 * 10^23

Which of the following is NOT a conversion factor for moles?

• 1 mole = 6.02 * 10^23 particles

• 1 mole = 22.4 L

• 1 mole = 28 g

• 1 mole = M mass in g

In the first sample problem, what is the volume of nitrogen gas given?

• 14.0 L

• 6.02 L

• 22.4 L

• 5.87 L

What is the atomic mass of nitrogen used in the first sample problem?

• 12

• 14

• 16

• 28

What is the mass of nitrogen gas calculated in the first sample problem?

• 7.34 g

• 5.87 g

• 22.4 g

• 28 g

What is the formula for nitrogen gas as mentioned in the first sample problem?

• NH3

• NO2

• N2

• N

In the second sample problem, what is the given mass of carbon dioxide?

• 44 g

• 12 g

• 32 g

• 58.2 g

What is the molar mass of carbon dioxide used in the second sample problem?

• 32 g/mol

• 44 g/mol

• 58.2 g/mol

• 28 g/mol

How many molecules of carbon dioxide are present in the given sample?

• 4.44 * 10^23

• 5.87 * 10^23

• 7.97 * 10^23

• 6.02 * 10^23

What is the atomic mass of oxygen used in the second sample problem?

• 12

• 14

• 18

• 16