0.3605 g of a metal is deposited on the electrode by passing 1.2 ampere current for 15 minutes through its salt. Atomic weight of the metal is 96. what will be its valency?

Explanation

To solve the problem step by step, we will follow the process of calculating the valency of the metal deposited on the electrode.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of metal deposited (W) = 0.3605 g
- Current (I) = 1.2 A
- Time (T) = 15 minutes
- Atomic weight of the metal = 96 g/mol

2. **Convert Time from Minutes to Seconds:**
\[ T = 15 \text{ minutes} \times 60 \text{ seconds/minute} = 900 \text{ seconds} \]

3. **Use the Formula for Mass Deposited:**
The formula to relate the mass deposited (W), electrochemical equivalence (Z), current (I), and time (T) is:
\[ W = Z \times I \times T \]

4. **Calculate the Electrochemical Equivalence (Z):**
First, we need to express Z in terms of equivalent weight (E) and Faraday's constant (F):
\[ Z = \frac{E}{F} \]
where \( F = 96000 \, \text{C/mol} \) (Faraday's constant).

5. **Rearranging the Formula to Find Equivalent Weight (E):**
From the mass deposited formula, we can rearrange it to find E:
\[ E = \frac{W \times F}{I \times T} \]

6. **Substituting the Values:**
Now, substituting the known values into the equation:
\[ E = \frac{0.3605 \, \text{g} \times 96000 \, \text{C/mol}}{1.2 \, \text{A} \times 900 \, \text{s}} \]

7. **Calculating Equivalent Weight (E):**
\[ E = \frac{0.3605 \times 96000}{1.2 \times 900} \]
\[ E = \frac{34608}{1080} \approx 32.2 \, \text{g/mol} \]

8. **Calculate the Valency (n):**
The valency (n) can be calculated using the formula:
\[ n = \frac{\text{Atomic Weight}}{\text{Equivalent Weight}} \]
Substituting the values:
\[ n = \frac{96 \, \text{g/mol}}{32.2 \, \text{g/mol}} \approx 2.98 \approx 3 \]

9. **Conclusion:**
The valency of the metal is approximately 3.