Alum is found to contain hydrated monovalent cation `[M(H_(2)O)_(6)]^(+)` trivalent cation `[M'(H_(2)O)_(6)]^(+)` and `SO_(4)^(3-)` in the ratio of:

Options

A. `1:1:1`
B. `1:2:3`
C. `1:3:2`
D. `1:1:2`

Explanation

To solve the problem regarding the composition of alum, we need to analyze the components given in the question. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Components of Alum
Alum is a type of double salt that typically contains:
- A hydrated monovalent cation, represented as `[M(H₂O)₆]⁺`
- A trivalent cation, represented as `[M'(H₂O)₆]³⁺`
- The sulfate ion, represented as `SO₄²⁻`

### Step 2: Write the General Formula
The general formula for alum can be expressed as:
\[ M_2SO_4 \cdot M'_{2(SO_4)} \cdot 24H_2O \]
This indicates that there are two monovalent cations (M), two trivalent cations (M'), and sulfate ions.

### Step 3: Identify the Ratio of Ions
From the general formula:
- The number of monovalent cations (M) = 2
- The number of trivalent cations (M') = 2
- The number of sulfate ions (SO₄²⁻) = 4

### Step 4: Simplify the Ratio
Now, we can express the ratio of the cations and sulfate ions:
- Ratio of M : M' : SO₄ = 2 : 2 : 4

To simplify this ratio, we divide each term by 2:
- M : M' : SO₄ = 1 : 1 : 2

### Step 5: Conclusion
Thus, the final ratio of the hydrated monovalent cation, trivalent cation, and sulfate ion in alum is:
\[ \text{Ratio} = 1 : 1 : 2 \]

### Final Answer
The ratio of hydrated monovalent cation `[M(H₂O)₆]⁺`, trivalent cation `[M'(H₂O)₆]³⁺`, and sulfate ion `SO₄²⁻` in alum is **1 : 1 : 2**.

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