What volume would 3.01•1023 molecules of oxygen gas occupy at stp? 11.2 ml 22.4 ml 11.2 l 22.4 l

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

 Volume  = 11.2 L

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error