X forms an oxide `X_(2)O_(3)` 0.36 grams of X forms 0.56 grams of `X_(2)O_(3)`. So the atomic weight of X is

Options

A. 36
B. 565
C. 28
D. 43.2

Answer

To find the atomic weight of element X that forms the oxide \( X_2O_3 \), we can follow these steps:

### Step 1: Write the reaction equation
The reaction for the formation of the oxide can be represented as:
\[ 4X + 3O_2 \rightarrow 2X_2O_3 \]
This indicates that 4 moles of element X react with 3 moles of oxygen to form 2 moles of the oxide \( X_2O_3 \).

### Step 2: Determine the molar mass of \( X_2O_3 \)
The molar mass of \( X_2O_3 \) can be expressed in terms of the atomic weight of X (let's denote it as \( m \)):
\[ \text{Molar mass of } X_2O_3 = 2m + 3 \times 16 = 2m + 48 \]
Here, we use \( 16 \) g/mol as the atomic weight of oxygen.

### Step 3: Set up the mass relationship
From the problem, we know that 0.36 grams of X produces 0.56 grams of \( X_2O_3 \). We can set up a proportion based on the molar masses:
\[ \frac{0.36 \text{ g of } X}{4m} = \frac{0.56 \text{ g of } X_2O_3}{2m + 48} \]

### Step 4: Cross-multiply to solve for \( m \)
Cross-multiplying gives us:
\[ 0.36(2m + 48) = 0.56(4m) \]
Expanding both sides:
\[ 0.72m + 17.28 = 2.24m \]

### Step 5: Rearrange the equation
Rearranging the equation to isolate \( m \):
\[ 2.24m - 0.72m = 17.28 \]
\[ 1.52m = 17.28 \]

### Step 6: Solve for \( m \)
Now, divide both sides by 1.52:
\[ m = \frac{17.28}{1.52} \approx 11.368 \text{ g/mol} \]

### Step 7: Round to the nearest whole number
Since atomic weights are typically expressed as whole numbers, we round 11.368 to 11.

### Conclusion
The atomic weight of element X is approximately **11 g/mol**.

---