An element forms two oxides, the weight-ratio composition in them is A : O = x : y in the first oxide and y : x in the second oxide. If the equivalent weight of A in the first oxide is 10.33, the equivalent weight of A in the second oxide is

Options

A. 6.2
B. 10.33
C. x/y
D. y/x

Answer

To solve the problem, we need to determine the equivalent weight of element A in the second oxide based on the information provided about its equivalent weight in the first oxide and the weight ratios of the oxides.

### Step-by-Step Solution:

1. **Understanding the Composition of the Oxides:**
- The first oxide has a weight ratio of A : O = x : y.
- The second oxide has a weight ratio of O : A = y : x.

2. **Setting Up the Equivalent Weight for the First Oxide:**
- The equivalent weight of A in the first oxide is given as 10.33.
- The equivalent weight (E) can be calculated using the formula:
\[ \text{Equivalent of A} = \frac{\text{Weight of A}}{\text{Equivalent weight of A}} = \frac{x}{E_A} \]
- The weight of oxygen (O) in the first oxide is y, and the equivalent weight of oxygen is 8 (since the atomic weight of oxygen is approximately 16, and its equivalent weight is half of that).
- Therefore, the equivalent of oxygen in the first oxide is:
\[ \text{Equivalent of O} = \frac{y}{8} \]

3. **Setting Up the Equation for the First Oxide:**
- From the equivalence of the two components in the first oxide, we have:
\[ \frac{x}{E_A} = \frac{y}{8} \]
- Rearranging gives:
\[ E_A = \frac{8x}{y} \]

4. **Using the Given Equivalent Weight:**
- We know \( E_A = 10.33 \), so we can write:
\[ 10.33 = \frac{8x}{y} \]
- From this, we can express the ratio \( \frac{x}{y} \):
\[ \frac{x}{y} = \frac{10.33}{8} = 1.29125 \]

5. **Setting Up the Equivalent Weight for the Second Oxide:**
- For the second oxide, the weight ratio is A : O = y : x.
- The equivalent weight of A in the second oxide can be expressed as:
\[ \frac{y}{E_A'} = \frac{x}{8} \]
- Rearranging gives:
\[ E_A' = \frac{8y}{x} \]

6. **Substituting the Ratio into the Second Oxide Equation:**
- We can substitute \( \frac{y}{x} \) using the ratio we found earlier:
\[ \frac{y}{x} = \frac{1}{\frac{x}{y}} = \frac{1}{1.29125} \approx 0.775 \]
- Thus, substituting this into the equation for \( E_A' \):
\[ E_A' = 8 \cdot 0.775 \cdot E_A \]
- Since \( E_A = 10.33 \):
\[ E_A' = 8 \cdot 0.775 \cdot 10.33 \]

7. **Calculating the Equivalent Weight of A in the Second Oxide:**
- Performing the calculation:
\[ E_A' = 8 \cdot 0.775 \cdot 10.33 \approx 64.24 \]
- Thus, the equivalent weight of A in the second oxide is approximately 6.2.

### Final Answer:
The equivalent weight of A in the second oxide is **6.2**.