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 these steps:

### Step 1: Write the reaction and identify initial conditions
The reaction is given as:
\[ A(g) \rightleftharpoons B(g) + C(g) \]

Initially, we have 10 moles of \( A \). Therefore:
- Initial moles of \( A = 10 \)
- Initial moles of \( B = 0 \)
- Initial moles of \( C = 0 \)

### Step 2: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation of \( A \). At equilibrium, the moles of each component can be expressed as:
- Moles of \( A \) at equilibrium = \( 10(1 - \alpha) \)
- Moles of \( B \) at equilibrium = \( 10\alpha \)
- Moles of \( C \) at equilibrium = \( 10\alpha \)

### Step 3: Calculate the total moles at equilibrium
The total moles at equilibrium can be calculated 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 4: Calculate the average molecular weight of the equilibrium mixture
The average molecular weight \( M_{avg} \) of the equilibrium mixture is given by:
\[ M_{avg} = \frac{\text{Total mass}}{\text{Total moles}} \]
The total mass can be calculated as:
\[ \text{Total mass} = \text{Moles of } A \times M_A + \text{Moles of } B \times M_B + \text{Moles of } C \times M_C \]
Substituting the values:
\[ = (10(1 - \alpha) \times 100) + (10\alpha \times 60) + (10\alpha \times 40) \]
\[ = 1000(1 - \alpha) + 600\alpha + 400\alpha \]
\[ = 1000 - 1000\alpha + 1000\alpha = 1000 \]
Thus, the total mass is \( 1000 \) grams.

Now substituting into the average molecular weight formula:
\[ M_{avg} = \frac{1000}{10 + 10\alpha} = 83 \]

### Step 5: Solve for \( \alpha \)
Setting up the equation:
\[ \frac{1000}{10 + 10\alpha} = 83 \]
Cross-multiplying gives:
\[ 1000 = 83(10 + 10\alpha) \]
\[ 1000 = 830 + 830\alpha \]
\[ 1000 - 830 = 830\alpha \]
\[ 170 = 830\alpha \]
\[ \alpha = \frac{170}{830} = 0.2 \]

### Step 6: Calculate the number of moles of \( C \) at equilibrium
Using the value of \( \alpha \):
\[ \text{Moles of } C = 10\alpha = 10 \times 0.2 = 2 \]

### Final Answer
The number of moles of \( C \) present at equilibrium is \( 2 \).

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