600 ml of amixture of O3 and O2 weighs 1g at NTP Calculate the volume of ozone in the - Chemistry - Some Basic Concepts of Chemistry

Question

600 ml of amixture of O3 and O2 weighs 1g
at NTP Calculate the volume of ozone in the mixture

Surya · Accepted Answer

molecular mass of oxygen = 32
molecular mass of ozone = 48

let us say volume of O2 = x ml
hence volume of O3 = 600 - x ml (volume of oxygen + ozone=600
mL)
at STP,
wt of x ml O2 = 32*x/22400 gm
wt of 600 - x ml O3 = 48*(600 - x)/22400 gm

EQUATION:
32*x/22400 + 48*(600 - x)/22400 = 1 (wt of x mL oxygen=x
moles*atomic weight);
32x + 28800 - 48x = 22400
16x = 28800 - 22400 = 6400
x = 6400/16 = 400 mL

HENCE VOLUME OF O2 IN THE MIXTURE = 400 ml

600 ml of amixture of O3 and O2 weighs 1g at NTP. Calculate the volume of ozone in the mixture.

600 ml of amixture of O₃ and O₂ weighs 1g at NTP. Calculate the volume of ozone in the mixture.