CBSE Class 11-science Answered
The reaction is :
2O3 3 O2
2 vol. 3 vol.
Let the volume of ozone in the initial mixrure be x mL
The volume of oxygen in the initial mixture will be (150-x) mL
Now, from the reaction we know that
2 volumes of ozone on decomposition will give 3 volume of oxygen.
x mL of ozone on decomposition will give (3/2)x volume of oxygen.
Total volume of mixture after decomposition will be = (3/2)x + (150 - x) mL
(3/2)x + (150 - x) = 160
x = 20 mL
So,the volume of ozone in the initial mixrure be 20 mL
The volume of oxygen in the initial mixture will be (150-x) mL = 130 mL
To calculate the weight percentage of oxone in the original mixture:
22.4 L of O3 at STP will weigh 48 g
20 L of O3 at STP will weigh (48/ 22.4) x 20 g = 42.86 g
2.4 L of O2 at STP will weigh 32 g
130 L of O2 at STP will weigh (32/ 22.4) x 130 g = 185.71 g
Total mass of the mixture = 42.86 + 185.71 = 228.57 g
Weight percentage of ozone in the original mixture = (42.86/ 228.57) x 100 = 18.75%