Question

The number of ions formed when cuproammonium sulphate is dissolved in water is:
a) 1
b) 2
c) 4
d) Zero

Answer 1

option ''b'' is correct

Answer 2

Answer:

b) 2

Explanation:

As per the question,

The chemical formula of Cuproammonium Sulphate is,

Cuproammonium Sulphate = $[Cu(NH_{3})_{4}]SO_{4}$

When it is dissolved in water, it gives two type of ions.

Dissociation reaction with water is:

$[Cu(NH_{3})_{4}]SO_{4} \Rightarrow\ [Cu(NH_{3})_{4}]^{2+} +[SO_{4}^{2-}]$

Therefore,

The number of ions formed when cuproammonium sulphate is dissolved in water is $[Cu(NH_{3})_{4}]^{2+} \ and \ [SO_{4}^{2-}]$

Hence, the correct option is (b).