What volume of carbon monoxide will contain 3.02x10^23 molecules?

11.23 L.

Explanation:

Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).

Using cross multiplication:

1.0 mole → 6.022 x 10²³ molecules.

? mole → 3.02 x 10²³ molecules.

The no. of moles of CO = (3.02 x 10²³ molecules) (1.0 mole) / (6.022 x 10²³ molecules) = 0.5 mol. We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.

Using cross multiplication:

1.0 mole of CO occupies → 22.4 L.

0.5 mole of CO occupies →? L.

∴ The volume of CO contain 3.02 x 10²³ molecules = (22.4 L) (0.5 mole) / (1.0 mole) = 11.23 L.