If the aqueous solutions of the following salts are electrolysed for 1 hour with 10 ampere current, which solution will deposit the maximum mass of the metal at the cathode. The atomic weights are, Fe = 56, Zn = 65, Ag = 108, Hf = 178 and W = 184

Options

A. `ZnSO_(4)`
B. `FeCl_(3)`
C. `HfCl_(4)`
D. `AgNO_(3)`

Explanation

To determine which solution will deposit the maximum mass of metal at the cathode when electrolyzed for 1 hour with a 10 ampere current, we can use the following steps:

### Step 1: Calculate the total charge (Q) passed during electrolysis
The total charge can be calculated using the formula:
\[ Q = I \times t \]
Where:
- \( I \) = current in amperes (10 A)
- \( t \) = time in seconds (1 hour = 3600 seconds)

So,
\[ Q = 10 \, \text{A} \times 3600 \, \text{s} = 36000 \, \text{C} \]

### Step 2: Calculate the number of moles of electrons (n)
Using Faraday's law, the number of moles of electrons can be calculated as:
\[ n = \frac{Q}{F} \]
Where:
- \( F \) = Faraday's constant (approximately 96500 C/mol)

So,
\[ n = \frac{36000 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.373 \, \text{mol} \]

### Step 3: Determine the equivalent weight of each metal
The equivalent weight (E) of a metal can be calculated using the formula:
\[ E = \frac{\text{Atomic weight}}{n} \]
Where \( n \) is the number of electrons transferred per ion.

1. **For Zinc (Zn)**: Atomic weight = 65, n = 2
\[ E_{Zn} = \frac{65}{2} = 32.5 \]

2. **For Iron (Fe)**: Atomic weight = 56, n = 3
\[ E_{Fe} = \frac{56}{3} \approx 18.67 \]

3. **For Silver (Ag)**: Atomic weight = 108, n = 1
\[ E_{Ag} = \frac{108}{1} = 108 \]

4. **For Hafnium (Hf)**: Atomic weight = 178, n = 4
\[ E_{Hf} = \frac{178}{4} = 44.5 \]

5. **For Tungsten (W)**: Atomic weight = 184, n = 6
\[ E_{W} = \frac{184}{6} \approx 30.67 \]

### Step 4: Calculate the mass of metal deposited for each solution
Using the formula:
\[ \text{Mass} = n \times E \]
Where \( n \) is the number of moles of electrons calculated in Step 2.

1. **For Zn**:
\[ \text{Mass}_{Zn} = 0.373 \, \text{mol} \times 32.5 \, \text{g/mol} \approx 12.14 \, \text{g} \]

2. **For Fe**:
\[ \text{Mass}_{Fe} = 0.373 \, \text{mol} \times 18.67 \, \text{g/mol} \approx 6.95 \, \text{g} \]

3. **For Ag**:
\[ \text{Mass}_{Ag} = 0.373 \, \text{mol} \times 108 \, \text{g/mol} \approx 40.24 \, \text{g} \]

4. **For Hf**:
\[ \text{Mass}_{Hf} = 0.373 \, \text{mol} \times 44.5 \, \text{g/mol} \approx 16.59 \, \text{g} \]

5. **For W**:
\[ \text{Mass}_{W} = 0.373 \, \text{mol} \times 30.67 \, \text{g/mol} \approx 11.43 \, \text{g} \]

### Step 5: Compare the masses deposited
- Mass of Zn deposited = 12.14 g
- Mass of Fe deposited = 6.95 g
- Mass of Ag deposited = 40.24 g
- Mass of Hf deposited = 16.59 g
- Mass of W deposited = 11.43 g

### Conclusion
The maximum mass of metal deposited at the cathode will be from the silver (Ag) solution, which is approximately 40.24 g.

### Final Answer
**Silver (Ag) solution will deposit the maximum mass of metal at the cathode.**

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