For the reaction
`A(g)hArrB(g)+C(g)`
At `300K`, the average molecular weight of the equilibrium mixture is `83g//"mol"`. If atomic weight of `A, B` and `C` are 100, 60 and 40 gram/mol respectively, then the number of moles of `'C'` present at equilibrium in a reaction starting with 10 moles of `A` is

Explanation

To solve the problem step by step, we will follow the outlined process based on the information given in the question.

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

### Step 2: Initial Moles
Initially, we start with 10 moles of A, and no moles of B or C:
- Moles of A = 10
- Moles of B = 0
- Moles of C = 0

### Step 3: Define Degree of Dissociation
Let \( \alpha \) be the degree of dissociation of A. At equilibrium:
- Moles of A = \( 10(1 - \alpha) \)
- Moles of B = \( 10\alpha \)
- Moles of C = \( 10\alpha \)

### Step 4: Total Moles at Equilibrium
The total moles at equilibrium can be expressed as:
\[ \text{Total moles} = \text{Moles of A} + \text{Moles of B} + \text{Moles of C} = 10(1 - \alpha) + 10\alpha + 10\alpha = 10 + 10\alpha \]

### Step 5: Average Molecular Weight
The average molecular weight of the equilibrium mixture is given as 83 g/mol. The average molecular weight can be calculated using the formula:
\[ \text{Average Molecular Weight} = \frac{\text{Total mass of the mixture}}{\text{Total moles of the mixture}} \]

The total mass of the mixture is:
\[ \text{Total mass} = (10(1 - \alpha) \cdot 100) + (10\alpha \cdot 60) + (10\alpha \cdot 40) \]
\[ = 1000(1 - \alpha) + 600\alpha + 400\alpha = 1000 - 1000\alpha + 1000\alpha = 1000 \]

So, the average molecular weight equation becomes:
\[ 83 = \frac{1000}{10 + 10\alpha} \]

### Step 6: Solve for \( \alpha \)
Rearranging the equation gives:
\[ 83(10 + 10\alpha) = 1000 \]
\[ 830 + 830\alpha = 1000 \]
\[ 830\alpha = 1000 - 830 \]
\[ 830\alpha = 170 \]
\[ \alpha = \frac{170}{830} = \frac{17}{83} \]

### Step 7: Calculate Moles of C
Now, we can find the number of moles of C at equilibrium:
\[ \text{Moles of C} = 10\alpha = 10 \times \frac{17}{83} = \frac{170}{83} \approx 2.05 \]

### Step 8: Conclusion
Since we are looking for the number of moles of C present at equilibrium, we can round it to the nearest whole number:
\[ \text{Moles of C} \approx 2 \]

Thus, the number of moles of C present at equilibrium is **2 moles**.

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