Question

A student has two samples A and B. A fresh solution of $FeS{O}_4$ was added to both the samples. After shaking both the samples well, a compound X was added. It was seen that in sample A a brown ring was formed but in the sample, B no brown ring is formed. Which of the following options could be A, B and X?

• A. A= Sodium sulphate, B= Silver chloride and X= ammonia
• B. A= Sodium Nitrate, B= Silver chloride and X= sulphuric acid
• C. A= Sodium sulphate, B= Silver chloride and X= sulphuric acid
• D. A= Sodium Nitrate, B= Silver chloride and X= ammonia

Answer 1

The correct option is B

A= Sodium Nitrate, B= Silver chloride and X= sulphuric acid

The brown ring formation is due to the presence of nitrates. A common nitrate test, known as the brown ring test is used to indicate the presence of nitrates. It is performed by adding freshly prepared iron (II) sulphate ( $FeS{O}_4$) to a solution of a given sample, then concentrated sulphuric acid is slowly added along the sides of the test tube such that the acid forms a layer below the aqueous solution. If a brown layer is formed then the given salt is a nitrate. After reading the procedure for the test it is clear that compound X is sulphuric acid.

So the option A= Sodium Nitrate, B= Silver chloride and X= ammonia is wrong as X is not sulphuric acid in this option.

In the question we are given that sample A forms a brown ring which means it is a nitrate compound. Since sample B did not form a brown ring it cannot be a nitrate.

Therefore the option A= Sodium sulphate, B= Silver chloride and X= sulphuric acid cannot be the answer as sample A here is not a nitrate.

The only option that satisfies all these conditions is A= Sodium Nitrate, B= Silver chloride and X= sulphuric acid.