When the same amount of zinc is treated separately with excess of `H_(2)SO_(4)` and excess of `NaOH`, the ratio of volumes of `H_(2)` evolved is:

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. `9:4`

Explanation

To solve the problem, we need to analyze the reactions that occur when zinc (Zn) is treated with excess sulfuric acid (H₂SO₄) and excess sodium hydroxide (NaOH).

### Step-by-step Solution:

1. **Identify the reactions**:
- When zinc reacts with sulfuric acid, the reaction can be represented as:
\[ \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \uparrow \]
- When zinc reacts with sodium hydroxide, the reaction can be represented as:
\[ \text{Zn} + 2\text{NaOH} + 2\text{H}_2\text{O} \rightarrow \text{Na}_2\text{Zn(OH)}_4 + \text{H}_2 \uparrow \]

2. **Determine the moles of hydrogen gas evolved**:
- In the first reaction with sulfuric acid, 1 mole of zinc produces 1 mole of hydrogen gas (H₂).
- In the second reaction with sodium hydroxide, 1 mole of zinc also produces 1 mole of hydrogen gas (H₂).

3. **Calculate the volume of hydrogen gas produced**:
- According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Therefore, the volume of hydrogen gas produced in both reactions will be the same if the amount of zinc is the same.

4. **Establish the ratio**:
- Since both reactions produce the same amount of hydrogen gas, the ratio of the volumes of hydrogen evolved when zinc is treated with excess H₂SO₄ and excess NaOH is:
\[ \text{Volume of } H_2 \text{ from } H_2SO_4 : \text{Volume of } H_2 \text{ from } NaOH = 1 : 1 \]

### Final Answer:
The ratio of volumes of hydrogen evolved when the same amount of zinc is treated separately with excess H₂SO₄ and excess NaOH is **1:1**.