116 g of `A_(3)B_(4)` has 1.5 moles of A. Molecular weight of `A_(3)B_(4)` is

Options

A. 164
B. 232
C. 77
D. cannot be calculated

Answer

To find the molecular weight of \( A_3B_4 \), we can follow these steps:

### Step 1: Determine the number of moles of \( A_3B_4 \)
Given that 116 g of \( A_3B_4 \) contains 1.5 moles of \( A \), we can find out how many moles of \( A_3B_4 \) correspond to this amount of \( A \).

Since each mole of \( A_3B_4 \) contains 3 moles of \( A \):
\[ \text{Moles of } A_3B_4 = \frac{\text{Moles of } A}{3} = \frac{1.5}{3} = 0.5 \text{ moles} \]

### Step 2: Calculate the molecular weight of \( A_3B_4 \)
Now, we know that 0.5 moles of \( A_3B_4 \) weigh 116 g. To find the weight of 1 mole of \( A_3B_4 \), we can use the following calculation:
\[ \text{Weight of 1 mole of } A_3B_4 = \text{Weight of 0.5 moles} \times 2 = 116 \text{ g} \times 2 = 232 \text{ g} \]

### Conclusion
Thus, the molecular weight of \( A_3B_4 \) is 232 g.

### Final Answer
The molecular weight of \( A_3B_4 \) is **232 g**.

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