A gaseous mixture contains three gaseous `A,B` and `C` with a total number of moles of 10 and total pressure of `10 atm`. The partial pressure of `A` and `B` are `3 atm` and `1` atm respectively and if `C` has molecular weight of `2 g //mol`. Then, the weight of `C` present in the mixture will be `:`

Options

A. 8 g
B. 12 g
C. 3 g
D. 6 g

Explanation

To solve the problem step by step, we will use the information provided about the gaseous mixture and apply the ideal gas law and the concepts of partial pressure and mole fraction.

### Step 1: Understand the given data
- Total moles of the gas mixture (n_total) = 10 moles
- Total pressure (P_total) = 10 atm
- Partial pressure of gas A (P_A) = 3 atm
- Partial pressure of gas B (P_B) = 1 atm
- Molecular weight of gas C (M_C) = 2 g/mol

### Step 2: Calculate the partial pressure of gas C
Using Dalton's Law of Partial Pressures, we know that:
\[ P_A + P_B + P_C = P_{total} \]

Substituting the known values:
\[ 3 \, \text{atm} + 1 \, \text{atm} + P_C = 10 \, \text{atm} \]

Now, solve for \( P_C \):
\[ P_C = 10 \, \text{atm} - 3 \, \text{atm} - 1 \, \text{atm} = 6 \, \text{atm} \]

### Step 3: Calculate the number of moles of gases A, B, and C
Using the relationship between partial pressure and mole fraction:
\[ P_A = X_A \cdot P_{total} \]
\[ P_B = X_B \cdot P_{total} \]
\[ P_C = X_C \cdot P_{total} \]

Where \( X_A, X_B, X_C \) are the mole fractions of gases A, B, and C respectively.

From the equation:
\[ X_A = \frac{n_A}{n_{total}} \]
\[ X_B = \frac{n_B}{n_{total}} \]
\[ X_C = \frac{n_C}{n_{total}} \]

We can express the number of moles:
1. For gas A:
\[ P_A = X_A \cdot P_{total} \]
\[ 3 = \frac{n_A}{10} \cdot 10 \]
\[ n_A = 3 \]

2. For gas B:
\[ P_B = X_B \cdot P_{total} \]
\[ 1 = \frac{n_B}{10} \cdot 10 \]
\[ n_B = 1 \]

3. Now, using the total moles equation:
\[ n_A + n_B + n_C = 10 \]
\[ 3 + 1 + n_C = 10 \]
\[ n_C = 10 - 4 = 6 \]

### Step 4: Calculate the weight of gas C
Using the formula:
\[ \text{Weight} = \text{moles} \times \text{molecular weight} \]
\[ \text{Weight of C} = n_C \times M_C \]
\[ \text{Weight of C} = 6 \, \text{moles} \times 2 \, \text{g/mol} = 12 \, \text{grams} \]

### Final Answer
The weight of gas C present in the mixture is **12 grams**.

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