A sample of mixture of `A(g), B (g)` and `C(g)` under equilibrium has mean molecular mass equal to 80. The equilibrium is `:`
`A(g)hArr B(g)+C(g)`
If initially 4 mole of `'A'` gas is present then total number of mole at equilibrium is `:`
`[M_(A)=100,M_(B)=60,M_(C)=40]`

Options

A. 5
B. 2
C. 6
D. 4

Explanation

To solve the problem step by step, we will analyze the equilibrium reaction and the information provided.

### Step 1: Write the equilibrium reaction
The equilibrium reaction given is:
\[ A(g) \rightleftharpoons B(g) + C(g) \]

### Step 2: Set up initial conditions
Initially, we have:
- Moles of A = 4
- Moles of B = 0
- Moles of C = 0

### Step 3: Define changes at equilibrium
Let \( x \) be the number of moles of A that dissociate into B and C at equilibrium. Therefore, at equilibrium:
- Moles of A = \( 4 - x \)
- Moles of B = \( x \)
- Moles of C = \( x \)

### Step 4: Calculate total moles at equilibrium
The total number of moles at equilibrium can be expressed as:
\[ \text{Total moles} = (4 - x) + x + x = 4 + x \]

### Step 5: Use the mean molecular mass
The mean molecular mass of the mixture is given as 80. The mean molecular mass can be calculated using the formula:
\[ \text{Mean Molecular Mass} = \frac{\sum (\text{Number of moles} \times \text{Molar mass})}{\text{Total moles}} \]
Substituting the values, we have:
\[ 80 = \frac{(4 - x) \times 100 + x \times 60 + x \times 40}{4 + x} \]

### Step 6: Simplify the equation
Expanding the numerator:
\[ 80 = \frac{400 - 100x + 60x + 40x}{4 + x} \]
This simplifies to:
\[ 80 = \frac{400 - 100x + 100x}{4 + x} \]
Thus, we have:
\[ 80 = \frac{400}{4 + x} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 80(4 + x) = 400 \]
Expanding this:
\[ 320 + 80x = 400 \]
Subtracting 320 from both sides:
\[ 80x = 80 \]
Dividing by 80:
\[ x = 1 \]

### Step 8: Calculate total moles at equilibrium
Now substituting \( x \) back into the total moles equation:
\[ \text{Total moles} = 4 + x = 4 + 1 = 5 \]

### Final Answer
The total number of moles at equilibrium is **5**.

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