A gaseous mixture of three gases A, B and C has a pressure of `10` atm. The total number of moles of all the gases is `10`. If the partial pressure of `A` and `B` are `3.0` and `1.0` atm respectively and if `C` has a mol/wt. of `2.0`. What is the weight of `C` in `g` present in the mixture ?

Options

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

Explanation

To solve the problem step by step, we will use the information provided about the gaseous mixture of gases A, B, and C.

### Step 1: Understand the given data
- Total pressure of the mixture (P_total) = 10 atm
- Total number of moles (n_total) = 10 moles
- Partial pressure of gas A (P_A) = 3.0 atm
- Partial pressure of gas B (P_B) = 1.0 atm
- Molecular weight of gas C (M_C) = 2.0 g/mol

### Step 2: Calculate the partial pressure of gas C
Using Dalton's Law of Partial Pressures, we can find the partial pressure of gas C (P_C):
\[ P_C = P_{total} - (P_A + P_B) \]
Substituting the known values:
\[ P_C = 10 \, \text{atm} - (3.0 \, \text{atm} + 1.0 \, \text{atm}) = 10 \, \text{atm} - 4.0 \, \text{atm} = 6.0 \, \text{atm} \]

### Step 3: Calculate the mole fractions of gases A and B
Using the formula for mole fraction:
\[ X_A = \frac{P_A}{P_{total}} \quad \text{and} \quad X_B = \frac{P_B}{P_{total}} \]
Calculating the mole fraction of gas A:
\[ X_A = \frac{3.0 \, \text{atm}}{10 \, \text{atm}} = 0.3 \]
Calculating the mole fraction of gas B:
\[ X_B = \frac{1.0 \, \text{atm}}{10 \, \text{atm}} = 0.1 \]

### Step 4: Calculate the mole fraction of gas C
Using the relationship:
\[ X_C = 1 - (X_A + X_B) \]
Substituting the values:
\[ X_C = 1 - (0.3 + 0.1) = 1 - 0.4 = 0.6 \]

### Step 5: Calculate the number of moles of gas C
Using the total number of moles:
\[ n_C = X_C \times n_{total} \]
Substituting the values:
\[ n_C = 0.6 \times 10 = 6 \, \text{moles} \]

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

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

---