In a reaction at equilibrium, X moles of the reactant A decomposes to give `1` mole each of C and D. It has been found that the fraction of A decomposed at equilibrium is independent of initial concentration of A. Calculate X.

Explanation

To solve the problem step by step, let's break down the information given and derive the solution systematically.

### Step 1: Write the Reaction
The decomposition of reactant A can be represented as:
\[ A \rightarrow C + D \]
where \( x \) moles of A decompose to give 1 mole each of C and D.

### Step 2: Define Initial and Equilibrium Concentrations
Let:
- The initial concentration of A be \( C \) (in moles).
- The fraction of A that decomposes be \( \alpha \).

At equilibrium:
- The concentration of A will be \( C(1 - \alpha) \).
- The concentration of C will be \( \frac{C\alpha}{x} \) (since 1 mole of C is produced for every \( x \) moles of A).
- The concentration of D will also be \( \frac{C\alpha}{x} \).

### Step 3: Write the Expression for the Equilibrium Constant (\( K_c \))
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[C][D]}{[A]^x} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{C\alpha}{x}\right) \left(\frac{C\alpha}{x}\right)}{(C(1 - \alpha))^x} \]
This simplifies to:
\[ K_c = \frac{\frac{C^2\alpha^2}{x^2}}{C^x(1 - \alpha)^x} \]
\[ K_c = \frac{C^2\alpha^2}{x^2 C^x (1 - \alpha)^x} \]
\[ K_c = \frac{\alpha^2}{x^2 (1 - \alpha)^x} \]

### Step 4: Analyze the Independence from Initial Concentration
The problem states that the fraction of A decomposed at equilibrium is independent of the initial concentration of A. This implies that the equilibrium constant \( K_c \) must not depend on \( C \).

To achieve this, we need to choose \( x \) such that the concentration term cancels out.

### Step 5: Determine the Value of \( x \)
To make \( K_c \) independent of the concentration \( C \), we can set \( x = 2 \).

Substituting \( x = 2 \) into the equilibrium expression:
\[ K_c = \frac{\alpha^2}{2^2 (1 - \alpha)^2} \]
\[ K_c = \frac{\alpha^2}{4(1 - \alpha)^2} \]
This expression is now independent of \( C \).

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{2} \]