Question

From a container having 64g oxygen ,11.2 l oxygen gas at s.t.p and 6.022 *10^23 oxygen atoms are removed. find the mass of the oxygen gas left

Answer 1

Hello Dear.

Original Mass of the Oxygen = 64 g.
No. of atoms Removed = 6.022 × 10²³ atoms.
At. Mass of the Oxygen = 16 g.

Using the Formula,
 
 No. of atoms = (Mass/At. Mass) × Avogadro's number.
∴ 6.022 × 10²³ = (Mass/16) × 6.022 × 10²³
  ⇒ Mass = 16 g.

Also,
We know, 1 mole of Gas occupies the volume of 22.4 liter.
∴ 22.4 liter of the oxygen gas weighs 32 g (1 mole).
∴ 1 liter of the O₂ of the oxygen weighs 32/22.4 g.
∴ 11.2 liter of the oxygen weighs 16 grams.

Now, Total Mass of the Oxygen Removed = 16 + 16
 = 32 grams.

Mass of the Oxygen left = Total Mass of the Oxygen - Mass of the Oxygen Removed.
= 64 - 32
= 32 g.


Hence, the mass of the oxygen left in the container is 32 g.


Hope it helps.

Answer 2

Original Mass of the Oxygen = 64 g.

No. of atoms Removed = 6.022 × 10²³ atoms.

At. Mass of the Oxygen = 16 g.

Using the Formula,

 

 No. of atoms = (Mass/At. Mass) × Avogadro's number.

∴ 6.022 × 10²³ = (Mass/16) × 6.022 × 10²³

 ⇒ Mass = 16 g.

Also,

We know, 1 mole of Gas occupies the volume of 22.4 liter.

∴ 22.4 liter of the oxygen gas weighs 32 g (1 mole).

∴ 1 liter of the O₂ of the oxygen weighs 32/22.4 g.

∴ 11.2 liter of the oxygen weighs 16 grams.

Now, Total Mass of the Oxygen Removed = 16 + 16

 = 32 grams.

Mass of the Oxygen left = Total Mass of the Oxygen - Mass of the Oxygen Removed.

= 64 - 32

= 32 g.

Hence, the mass of the oxygen left in the container is 32 g