Question

Calculate the volume occupied of STP by (i) 16.0g of oxygen (ii) 1.5moles of oxygen (iii) 6.022 × 10^23 molecule of carbon dioxide​

Answer 1

Concept :- Volume occupied by one mole of gas at STP is 22.4 litre.

(I) Molar mass of O2 = 32 g.

Hence, number of moles of oxygen in 16 gram of oxygen is 0.5

Hence, volume occupied = no. of moles × 22.4 = 0.5 × 22.4 = 11.2 litre

(ii) Volume occupied = no. of moles × 22.4 == 1.5 × 22.4 = 33.6 litre.

(iii) 6.022 × 10^23 molecules = 1 moles.

Hence, volume occupied = 22.4 litre

Answer 2

Answer:

1) 11.2 L

2)33.6 L

3) 22.4 L

Explanation:

Any gas of one mole will occupy 22.4 L

hence

16g O2= 0.5 mole O2 = 11. 2 L

1.5 mole O2 = 33.6 L

6.022 x 10^ 23 molecule of CO2 = 1 mole of CO2 = 22.4 L