Preparation of Mohr's salt and potash alum — practice questions

Question 1 · difficulty L1 · recall

What type of compound is Mohr's salt, FeSO4·(NH4)2SO4·6H2O?

• A. A double salt
• B. A coordination compound
• C. A simple binary salt
• D. A covalent molecular solid

Question 2 · difficulty L2 · understanding

Why is a small amount of dilute sulphuric acid used while preparing crystals of Mohr's salt?

• A. It precipitates metallic iron from the solution
• B. It limits Fe2+ hydrolysis and oxidation
• C. It converts ammonium ions into ammonia gas
• D. It removes sulphate ions from the mixture

Question 3 · difficulty L3 · application

A hot concentrated solution used to prepare a double salt is allowed to cool slowly. What is the main purpose of this step?

• A. To evaporate all water instantaneously
• B. To decompose the salt into its component metals
• C. To crystallise the salt as the solution cools
• D. To convert the dissolved ions into gases

Question 4 · difficulty L3 · application

During crystal preparation, why are the separated crystals washed with only a small amount of cold washing liquid?

• A. To convert the double salt into a coordination compound
• B. To increase the solubility of the product as much as possible
• C. To melt the crystals before drying
• D. To clean crystals while minimising product loss

Question 5 · difficulty L4 · analysis

A student obtains very little potash alum after concentrating the solution but filters it while it is still hot. What is the most likely reason for the low yield?

• A. The alum was still dissolved in the hot solution
• B. The alum had already changed into metallic aluminium
• C. Hot filtration always destroys sulphate ions
• D. Potassium ions evaporated from the solution