Elements A and B form two compounds `B_2​A_3` and `B_2​A`. If 0.05 moles of `B_2​A_3` weigh 9.0 g and 0.10 mole of `B_2​A` weigh 10 g, then atomic weight of A and B are :-

Options

A. 20 and 30
B. 30 and 40
C. 40 and 30
D. 30 and 20

Explanation

To find the atomic weights of elements A and B from the given compounds \( B_2A_3 \) and \( B_2A \), we can follow these steps:

### Step 1: Calculate the molar mass of \( B_2A_3 \)

We know that:
- Moles of \( B_2A_3 = 0.05 \)
- Mass of \( B_2A_3 = 9.0 \, \text{g} \)

Using the formula for molar mass:
\[ \text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}} \]
Substituting the values:
\[ \text{Molar Mass of } B_2A_3 = \frac{9.0 \, \text{g}}{0.05 \, \text{mol}} = 180 \, \text{g/mol} \]

### Step 2: Calculate the molar mass of \( B_2A \)

We know that:
- Moles of \( B_2A = 0.10 \)
- Mass of \( B_2A = 10.0 \, \text{g} \)

Using the same formula for molar mass:
\[ \text{Molar Mass of } B_2A = \frac{10.0 \, \text{g}}{0.10 \, \text{mol}} = 100 \, \text{g/mol} \]

### Step 3: Set up equations based on the molar masses

From the molar masses, we can set up the following equations based on the molecular formulas:
1. For \( B_2A_3 \):
\[ 2B + 3A = 180 \quad \text{(Equation 1)} \]
2. For \( B_2A \):
\[ 2B + A = 100 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations

To eliminate \( B \), we can subtract Equation 2 from Equation 1:
\[ (2B + 3A) - (2B + A) = 180 - 100 \]
This simplifies to:
\[ 2A = 80 \]
Dividing both sides by 2 gives:
\[ A = 40 \]

### Step 5: Substitute \( A \) back to find \( B \)

Now, substitute \( A = 40 \) back into Equation 2:
\[ 2B + 40 = 100 \]
Subtracting 40 from both sides gives:
\[ 2B = 60 \]
Dividing both sides by 2 gives:
\[ B = 30 \]

### Final Answer

The atomic weights are:
- Atomic weight of A = 40
- Atomic weight of B = 30