Question

800 ml of a mixtures of o2 and o3 weight 1.2 gm at NTP.the volume of o3 in mixture​

Answer 1

Answer:

Let the volume of ozone be V ml. Volume of oxegen = (600-V) ml Mass of 22400 ml of ozone at STP = 48 g ( Molar mass of O3 is 48 ) Mass of V ml of ozone = (48 ...

Answer 2

Answer:

volume of o3 in the mixture is 80ml

Explanation:

let o3 volume in the mixture be X ml.

then o2 volume becomes 800-X ml.

moles in terms of mass:

moles of 02= mass of 02/molar mass of o2

mole of o2= mass of o2/32.............(1)

moles in terms of volume:

converting 22.4 liters=22400ml because volume is taken in ml

moles of o2 = 800-x/22400...........(2)

equation 1=equation 2......(since, moles of oxygen are equal)

mass of 02 = 32(800-x)/22400.........(3)

SIMILARLY,

mass of 03=48x/22400......(4)

mass of o2 + mass of o3 = mass of mixture

32(800-x)/22400+48x/22400=1.2g

by solving this equation x=80ml

Hence, volume of ozone (o3) is 80ml at NTP