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 find the atomic weights of elements X and Y, we will use the information given about the compounds X₂Y₃ and X₃Y₄.

### Step 1: Set up the equations based on the given information.

1. For the compound X₂Y₃:
- We know that 0.20 moles of X₂Y₃ weigh 32.0 g.
- The molar mass of X₂Y₃ can be calculated as:
\[ \text{Molar mass of } X₂Y₃ = \frac{\text{Total weight}}{\text{Number of moles}} = \frac{32.0 \, \text{g}}{0.20 \, \text{mol}} = 160 \, \text{g/mol} \]
- The molar mass can also be expressed in terms of the atomic weights of X and Y:
\[ \text{Molar mass of } X₂Y₃ = 2M_X + 3M_Y \]
- Therefore, we have our first equation:
\[ 2M_X + 3M_Y = 160 \quad \text{(Equation 1)} \]

2. For the compound X₃Y₄:
- We know that 0.40 moles of X₃Y₄ weigh 92.8 g.
- The molar mass of X₃Y₄ can be calculated as:
\[ \text{Molar mass of } X₃Y₄ = \frac{92.8 \, \text{g}}{0.40 \, \text{mol}} = 232 \, \text{g/mol} \]
- The molar mass can also be expressed in terms of the atomic weights of X and Y:
\[ \text{Molar mass of } X₃Y₄ = 3M_X + 4M_Y \]
- Therefore, we have our second equation:
\[ 3M_X + 4M_Y = 232 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations.

Now we will solve the two equations simultaneously.

1. From Equation 1:
\[ 2M_X + 3M_Y = 160 \quad \text{(1)} \]

2. From Equation 2:
\[ 3M_X + 4M_Y = 232 \quad \text{(2)} \]

We can multiply Equation 1 by 3 to align the coefficients of \(M_X\):
\[ 6M_X + 9M_Y = 480 \quad \text{(Equation 3)} \]

Now we multiply Equation 2 by 2:
\[ 6M_X + 8M_Y = 464 \quad \text{(Equation 4)} \]

### Step 3: Subtract Equation 4 from Equation 3.

Subtracting Equation 4 from Equation 3 gives:
\[ (6M_X + 9M_Y) - (6M_X + 8M_Y) = 480 - 464 \]
This simplifies to:
\[ M_Y = 16 \]

### Step 4: Substitute \(M_Y\) back into one of the equations to find \(M_X\).

Now, substitute \(M_Y = 16\) into Equation 1:
\[ 2M_X + 3(16) = 160 \]
This simplifies to:
\[ 2M_X + 48 = 160 \]
\[ 2M_X = 160 - 48 \]
\[ 2M_X = 112 \]
\[ M_X = 56 \]

### Conclusion

The atomic weights of elements X and Y are:
- \(M_X = 56\)
- \(M_Y = 16\)