X and Y are two elements which form `X_(2)Y_(3)` and `X_(3)Y_(4)`. If `0.20` mol of `X_(2)Y_(3)` weight `32.0g` and `0.4mol` of `X_(3)Y_(4)` weigths `92.8g`, the atomic weights of X and Y are respectively

Options

A. `16.0` and `56.0`
B. `8.0` and `28.0`
C. `56.0` and `16.0`
D. `28.0` and `8.0`

Explanation

To solve the problem, we need to find the atomic weights of elements X and Y based on the given compounds and their respective weights and moles.

### Step 1: Write the equations for the compounds
We have two compounds:
1. \( X_2Y_3 \)
2. \( X_3Y_4 \)

### Step 2: Set up the equations based on the given data
From the problem, we know:
- For \( X_2Y_3 \):
- Moles = 0.20
- Weight = 32.0 g
- For \( X_3Y_4 \):
- Moles = 0.40
- Weight = 92.8 g

Using the formula:
\[ \text{Weight} = \text{Moles} \times \text{Molar Mass} \]

We can derive the molar mass for each compound:
1. For \( X_2Y_3 \):
\[ 32.0 = 0.20 \times (2x + 3y) \quad \text{(Equation 1)} \]
Rearranging gives:
\[ 2x + 3y = \frac{32.0}{0.20} = 160 \quad \text{(Equation 1)} \]

2. For \( X_3Y_4 \):
\[ 92.8 = 0.40 \times (3x + 4y) \quad \text{(Equation 2)} \]
Rearranging gives:
\[ 3x + 4y = \frac{92.8}{0.40} = 232 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have the following system of equations:
1. \( 2x + 3y = 160 \) (Equation 1)
2. \( 3x + 4y = 232 \) (Equation 2)

We can solve these equations simultaneously.

#### Step 3.1: Multiply Equation 1 by 3
\[ 3(2x + 3y) = 3(160) \]
\[ 6x + 9y = 480 \quad \text{(Equation 3)} \]

#### Step 3.2: Multiply Equation 2 by 2
\[ 2(3x + 4y) = 2(232) \]
\[ 6x + 8y = 464 \quad \text{(Equation 4)} \]

#### Step 3.3: Subtract Equation 4 from Equation 3
\[ (6x + 9y) - (6x + 8y) = 480 - 464 \]
\[ y = 16 \]

### Step 4: Substitute \( y \) back to find \( x \)
Now substitute \( y = 16 \) back into Equation 1:
\[ 2x + 3(16) = 160 \]
\[ 2x + 48 = 160 \]
\[ 2x = 160 - 48 \]
\[ 2x = 112 \]
\[ x = 56 \]

### Final Result
The atomic weights of \( X \) and \( Y \) are:
- Atomic weight of \( X = 56 \)
- Atomic weight of \( Y = 16 \)