1.60 g of a metal A and 0.96 g of a metal B when treated with excess of dilute acid, separately, produced the same amount of hydrogen. Calculate the equivalent weight of A if the equivalent weight of B is 12.

Answer

To solve the problem, we need to calculate the equivalent weight of metal A given that 1.60 g of metal A and 0.96 g of metal B produce the same amount of hydrogen when treated with excess dilute acid. We know the equivalent weight of metal B is 12.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- When metals A and B react with dilute acid, they produce hydrogen gas (H₂). The reactions can be represented as:
- A + H⁺ → H₂
- B + H⁺ → H₂

2. **Let the Amount of Hydrogen Produced be X grams**:
- We will denote the amount of hydrogen produced by both metals as X grams.

3. **Using the Concept of Equivalent Weight**:
- The equivalent weight (E) of a substance can be defined as:
\[ E = \frac{\text{Molecular Weight}}{\text{n factor}} \]
- The equivalent weight of metal B is given as 12. Therefore:
\[ E_B = \frac{M_B}{n_B} = 12 \]
- Rearranging gives us:
\[ M_B = 12 \times n_B \]

4. **Setting Up the Equivalence for Metal B**:
- The equivalent of metal B can be expressed as:
\[ \text{Equivalence of B} = \frac{0.96 \, \text{g}}{M_B} \times n_B \]
- Since both metals produce the same amount of hydrogen, we have:
\[ \frac{0.96}{M_B} \times n_B = \frac{X}{2} \]

5. **Setting Up the Equivalence for Metal A**:
- Similarly, for metal A:
\[ \text{Equivalence of A} = \frac{1.60 \, \text{g}}{M_A} \times n_A \]
- This gives us:
\[ \frac{1.60}{M_A} \times n_A = \frac{X}{2} \]

6. **Equating the Two Expressions**:
- Since both expressions equal X/2, we can set them equal to each other:
\[ \frac{0.96}{M_B} \times n_B = \frac{1.60}{M_A} \times n_A \]

7. **Substituting for M_B**:
- Substitute \( M_B = 12 \times n_B \) into the equation:
\[ \frac{0.96}{12 \times n_B} \times n_B = \frac{1.60}{M_A} \times n_A \]
- Simplifying gives:
\[ \frac{0.96}{12} = \frac{1.60}{M_A} \times n_A \]

8. **Calculating the Amount of Hydrogen (X)**:
- From the equivalence of B:
\[ \frac{0.96}{12} = \frac{1.60}{M_A} \times n_A \]
- Rearranging gives:
\[ M_A = \frac{1.60 \times 12}{0.96} \times n_A \]

9. **Finding the Equivalent Weight of Metal A**:
- The equivalent weight of A can be expressed as:
\[ E_A = \frac{M_A}{n_A} \]
- Substituting \( M_A \) gives:
\[ E_A = \frac{\frac{1.60 \times 12}{0.96} \times n_A}{n_A} = \frac{1.60 \times 12}{0.96} \]
- Simplifying this:
\[ E_A = \frac{19.2}{0.96} = 20 \]

### Final Answer:
The equivalent weight of metal A is **20 g/equiv**.