Question

22g of CO2 at STP occupies a volume​

Answer 1

Answer:

no. of mole of CO2 = Volume of CO2/22.4 in litres

One mole of CO2 = 12 + 16X2

= 44 g

Therefore, no. of mole of CO2 in 22 g = 22/44 = 0.5 mole

Therefore,

0.5 = V/22.4

V = 22.4 X 0.5 = 11.2 litres Ans.

Explanation:

Answer 2

Given :

Mass of CO₂ = 22 g

To Find :

Volume occupied by 22 g of CO₂ at STP

Solution :

We know, 1 mole of any gaseous substance at STP occupies a volume of 22.4 L.

We are required to find the volume occupied by 22 g of CO₂

As 44 g of CO₂ = 1 mole. So,

44 g of CO₂ occupies a volume of 22.4 L at STP

22 g of CO₂ will occupy a volume $\frac{22.4 }{44}$ × 22 L at STP

                                                     =  11.2 L at STP.

Therefore, volume occupied by 22 g of CO₂ at STP is 11.2 L.