Two voltameters, one having a solution of silver salt and the other of a trivalent-metal salt, are connected in series and a current of `2 A` is maintained for `1.50 hours`. It is found that `1.00 g` of the trivalent metal is deposited. (a) What is the atomic weight of the trivalent metal? (b) How much silver is deposited during this period?
Atomic weight of silver is `109.9 g mol^(-1)`

Explanation

To solve the problem step by step, we will break it down into two parts as requested: (a) finding the atomic weight of the trivalent metal, and (b) determining how much silver is deposited during the given time.

### Step-by-Step Solution

**(a) Finding the Atomic Weight of the Trivalent Metal:**

1. **Identify Given Values:**
- Current (I) = 2 A
- Time (t) = 1.5 hours = 1.5 × 3600 seconds = 5400 seconds
- Mass of trivalent metal deposited (m) = 1.00 g

2. **Use Faraday's Law of Electrolysis:**
\[ m = z \cdot I \cdot t \]
where \( z \) is the electrochemical equivalent.

3. **Calculate Electrochemical Equivalent (z):**
Rearranging the formula gives:
\[ z = \frac{m}{I \cdot t} \]
Substituting the known values:
\[ z = \frac{1.00 \, \text{g}}{2 \, \text{A} \cdot 5400 \, \text{s}} = \frac{1.00}{10800} \approx 0.0000926 \, \text{g/C} \]

4. **Determine Equivalent Weight:**
The equivalent weight (E) of the trivalent metal can be calculated using:
\[ E = \frac{z \cdot F}{n} \]
where \( F \) (Faraday's constant) = 96500 C/mol and \( n \) (valency) = 3 (since it's trivalent).
\[ E = \frac{0.0000926 \cdot 96500}{3} \approx 2.97 \, \text{g/mol} \]

5. **Calculate Molar Mass (M):**
The molar mass (M) is given by:
\[ M = E \cdot n = 2.97 \cdot 3 \approx 8.91 \, \text{g/mol} \]

6. **Final Result for Atomic Weight:**
The atomic weight of the trivalent metal is approximately **8.91 g/mol**.

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**(b) Finding the Mass of Silver Deposited:**

1. **Identify the Electrochemical Equivalent of Silver (z2):**
The atomic weight of silver (Ag) is given as 109.9 g/mol. Thus, the electrochemical equivalent of silver (z2) can be calculated as:
\[ z_2 = \frac{109.9}{96500} \approx 0.00114 \, \text{g/C} \]

2. **Use the Relationship Between Masses:**
From the relationship between the masses deposited:
\[ \frac{m_1}{m_2} = \frac{z_1}{z_2} \]
where \( m_1 \) is the mass of the trivalent metal and \( m_2 \) is the mass of silver deposited.

3. **Rearranging to Find m2:**
\[ m_2 = m_1 \cdot \frac{z_2}{z_1} \]
Substituting the values:
\[ m_2 = 1.00 \cdot \frac{0.00114}{0.0000926} \approx 12.3 \, \text{g} \]

4. **Final Result for Mass of Silver Deposited:**
The mass of silver deposited during this period is approximately **12.3 g**.

### Summary of Results:
- (a) Atomic weight of the trivalent metal: **8.91 g/mol**
- (b) Mass of silver deposited: **12.3 g**

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