`C(s)hArr2A(g)+B(s)`
If the dissociation of `C(s)` is `alpha` and `d` is the density of the gaseous mixture in the container. Initially container have only `C(s)` and the reaction is carried at constant temperature and pressure.

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the dissociation of solid carbon (C) into gaseous A and solid B, considering the dissociation fraction (α) and the density (D) of the gaseous mixture.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction is given as:
\[ C(s) \rightleftharpoons 2A(g) + B(s) \]
Here, C is in solid state, A is in gaseous state, and B is also in solid state.

2. **Initial Conditions**:
Initially, the container only contains solid C. Let’s denote the initial amount of C as \( n_0 \).

3. **Dissociation of C**:
If the dissociation of C is α, then the amount of C that dissociates is \( \alpha n_0 \). Therefore, the amount of C remaining after dissociation is:
\[ n_C = n_0 - \alpha n_0 = n_0(1 - \alpha) \]

4. **Formation of A and B**:
From the stoichiometry of the reaction, for every mole of C that dissociates, 2 moles of A are produced. Thus, the moles of A produced are:
\[ n_A = 2\alpha n_0 \]
The amount of solid B formed is equal to the amount of C that dissociated:
\[ n_B = \alpha n_0 \]

5. **Total Moles in the Gaseous Mixture**:
Since B is solid and does not contribute to the gas phase, the total moles in the gaseous phase is just the moles of A:
\[ n_{gas} = n_A = 2\alpha n_0 \]

6. **Density of the Gaseous Mixture**:
The density (D) of the gaseous mixture is given by the formula:
\[ D = \frac{m_{gas}}{V} \]
where \( m_{gas} \) is the mass of the gas and \( V \) is the volume. The mass of the gas is:
\[ m_{gas} = n_A \times M_A = 2\alpha n_0 M_A \]
Thus, the density can be expressed as:
\[ D = \frac{2\alpha n_0 M_A}{V} \]

7. **Conclusion on Density**:
Since the reaction is carried out at constant temperature and pressure, the volume (V) can be expressed using the ideal gas law:
\[ PV = n_{gas}RT \]
Substituting \( n_{gas} \):
\[ V = \frac{2\alpha n_0 RT}{P} \]
Now substituting V back into the density equation gives:
\[ D = \frac{2\alpha n_0 M_A P}{2\alpha n_0 RT} = \frac{M_A P}{RT} \]
This shows that the density D is independent of α, as it remains constant regardless of the dissociation of C.

### Final Answer:
The density of the gaseous mixture remains constant as α increases.