Question

Number of water molecules in Mohr’s salt is:
A. $7$
B. $6$
C. $5$
D. $8$

Answer 1

Number of water molecules present in a salt is also called as the water of crystallization. Hence the water of crystallization is the fixed number of water molecules which are attached chemically with a salt in crystalline form. Here Mohr’s salt is nothing but Ammonium iron sulphate and also is a double salt. A double salt is a salt which contains more than one anion and cation in it.

Complete answer:
Mohr’s salt is known as a double salt as when it is dissolved in water it undergoes complete dissociation and dissociates into $F{e^{2 + }},N{H_4}^ + ,S{O_4}^{2 - }$ and will then give separate chemical tests for each ion present.

Preparation of Mohr’s Salt: To prepare Mohr’s salt we need to dissolve equimolar mixture of hydrated ferrous sulphate ammonium sulphate in water with a little amount of sulphuric acid dissolved in water. After this we will proceed with the process of crystallization. At the end of the process we will obtain light green crystals of ferrous ammonium sulphate. The crystals prepared will be octahedral in shape.
The chemical reaction for the preparation of Mohr’s salt is:
$FeS{O_4} + {(N{H_4})_2}S{O_4} + 6{H_2}O \to FeS{O_4}.{(N{H_4})_2}S{O_4}.6{H_2}O$
So here we see that the number of water molecules present in Mohr’s salt will be $6$.

**Hence the correct answer is Option B.

Note:**
A little amount of sulphuric acid is added in the preparation of Mohr’s salt because the ferrous ions which are present in Mohr’s salt undergo hydrolysis when in aqueous solution. To prevent this concentrated sulphuric acid is added. Also, another reason is to prevent the oxidation of $F{e^{2 + }}$ ions of Mohr’s salt to $F{e^{3 + }}$ ions.