Question

What volume would 3.01•1023 molecules of oxygen gas occupy at STP?

Answer 1

First, find moles of oxygen gas:
(3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2

Second, multiply moles by the standard molar volume of a gas at STP:
(0.5mol)(22.4L) = 11.2L O2

Answer 2

Answer:

                  Volume  = 11.2 L

Explanation:

Step 1: Calculate Moles:

                      As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                          Number of Moles  =  Number of Particles ÷ 6.022 × 10²³

Putting values,

                          Number of Moles  =  3.01× 10²³ Particles ÷ 6.022 × 10²³

                          Number of Moles  =  0.50 Moles

Step 2: Calculate Volume:

As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ volume.

When 1 mole gas occupies 22.4 dm³ at STP then the volume occupied by 0.50 moles of gas is calculated as,

                      = (22.4 dm³ × 0.50 moles) ÷ 1 mole

                      = 11.2 dm³                                       ∴ 1dm³ = 1 L

So,

                                      Volume  = 11.2 L