Question

Calculate the mass and volume and mass of 0.2 mole of O3 at STP​

Answer 1

Explanation:

$\bold{(a) \: \sf{For \: Calculation \: of \: mass}} \\ \sf{Mass \: of \: 1 \: mole \: of \: O_{3} = 48g} \\ \sf{Mass \: of \: 0.2 \: mole \: of \: O_{3} = 48 \times 0.2gm} \\ \boxed{ \sf{Mass \: of \: 0.2 \: mole \: of \: O_{3} = 9.6g}} \\ \\ \bold{(b) \: \sf{For \: Calculation \: of \: volume}} \\ \sf{Volume \: of \: 1 \: mole \: of \: O_{3} \: at \: STP = 22.4L} \\ \sf{Volume \: of \: 0.2 \: mole \: of \: O_{3} \: at \: STP = 22.4 \times 0.2} \\ \boxed{\sf{Volume \: of \: 0.2 \: mole \: of \: O_{3} \: at \: STP = 4.48 \: litre}}$[/tex]

Answer 2

Answer:

Given,

• the number of moles (n) = 0.2
• the rmm of O3 = 16+ 16+16 = 48 amu

the number of moles = mass / rmm

=> 0.2 = mass / 48

mass= 9.6 gm.

• the number of moles (n) = 0.2
• the volume= V

V = n× 22.4

V = 0.2× 22.4 = 4.48 lts at s.t.p