# Decoding the Mystery- Equivalent mass of ozone/ n-factor of ozone/Reaction btw PbS and O3

### Problem 1 // Correct reaction b/w  PbS and O3 is represented by

(a) PbS + 2O3 = PbSO4 + O2

(b) PbS + 4O3 = PbSO4 + 4O2

(c) PbS + 6O3 = PbSO4 + 7O2

(d) 3PbS + 4O3 = 3PbSO4

### Problem 2 // Equivalent mass of ozone is

(a) 48

(b) 24

(c) 16

(d) 8

Problem 3 // How many mole KI(aq.) reacts with One mole O3 ?

(a) 1

(b) 2

(c) 3

(d) 4

### Discussion/Detailed Solution

1. During reaction, One Ozone molecule gives one O2 molecule and one nascent Oxygen atom, [O]. This nascent oxygen atom gains “two” electrons and becomes “oxide ion” and gets attached to the other substance. Hence
2. ### n-factor of ozone is 2 and its equivalent mass = 48/2 = 24

3. To balance the reaction b/w PbS and Ozone, there can be answer like, PbS + 2O3 = PbSO4 + O2 and nothing seems wrong at first sight.
4. One Ozone molecule really only gives one oxygen atom to the substance it oxidises.
5. Here, we must need  4 O3 molecules to provide “four oxygen atom” to convert one PbS molecule into one molecule of PbSO4
6. In other words, n-factor of O3 is 2 and “n-factor of PbS is 8” (change in oxidation number of sulfur from -2 to +6). Hence “mole ratio of PbS/O3 must be 2/8 = ¼ .”
7. Hence, correct answer is PbS + 4O3 = PbSO4 + 4O2
8. Similarly, n-factor of KI is 1 when it converts into I2, hence mole ratio KI/O3 = (2/1).
9. 2KI + H2O + O3 = 2KOH + I2 + O2
10. Hence answer of each of the above questions is option (B)