`CuSO_(4).5H_(2)O` becomes anhydrous `CuSO_(4)` on heating. This property of losing water molecules of crystallisation is known as

Options

A. deliquescence
B. dehydration
C. efflorescence
D. hydrolysis

Answer

To solve the question regarding the property of `CuSO4.5H2O` losing its water molecules of crystallization upon heating, we can break it down into the following steps:

### Step 1: Understand the Composition of `CuSO4.5H2O`
- `CuSO4.5H2O` is a hydrated form of copper(II) sulfate, which contains five water molecules associated with each formula unit of copper(II) sulfate.

### Step 2: Heating the Hydrated Salt
- When `CuSO4.5H2O` is heated, it undergoes a physical change where it loses the water molecules. This process occurs because the heat provides enough energy to break the bonds between the water molecules and the copper sulfate.

### Step 3: Formation of Anhydrous `CuSO4`
- After heating, the compound transforms into anhydrous `CuSO4`, which is the form of copper(II) sulfate without water.

### Step 4: Identify the Property
- The ability of a hydrated compound to lose water molecules upon heating is a specific property. This phenomenon is known as **dehydration**.

### Conclusion
- Therefore, the property of losing water molecules of crystallization when `CuSO4.5H2O` is heated is called **dehydration**.

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