 # 600 mL of a mixture of Ozone(O3) and Oxygen(O2) weighs 1 g at NTP.Calculate the volume of Ozone(O3)in the mixture.

10 years ago

Assume xg of Ozone . O2= 1-x g

(Total moles of gas)/22.4 = 0.6 L

{x/48 + (1-x)/32} / 22.4 = 0.6

Solve for x...

5 years ago
Hey x will give the amount of O3 to find the volume of O3 we need to apply that V/22.4 = no. of moles = x/48
Here x=3/7 g
therefore :-
V=0.2 Lt.
4 years ago
32g of O2= 22.4l
1g=22.4/32 l
xg= 22.4x/32

Similarly, (1-x)g O3=(1-x)22.4/48 litres

Sum=>        [22.4/x +(1-x)22.4/48] litres = 0.6 litres
Solving, x= 4/7 g Oxygen
(1-x)= 3/7 g Ozone
ie. 3/7 ×22.4/48 litres = 0.2 litres
=200 ml

3 years ago

Let the volume of ozone be V ml.
Volume of oxegen = (600-V) ml
Mass of 22,400 ml of ozone at STP = 48 g (Molar mass of O3 is 48 )
Mass of V ml of ozone = (48 x V/22400) g
Mass of 22,400 ml of oxygen = 32 g (Molar mass of O2 is 32)
Mass of (600-V) ml of oxygen = [32 x (600-V)}g]/22400
Now, [ 48 x V/22400 + 32 x (600-V)/22400]g = 1 g
[48 x V + 32 x (600-v)]/22400 = 1

48 x V + 32 x 600 - 32 V = 22400
48 V- 32 V = 22400 - (32 x 600)
16 V = 3200
V = 200 ml.