A certain metal salt solution is electrolysed in series with a silver coulometer. The weight of silver and the metal deposited are `0.5094g` and `0.2653g`. Calculate the valency of the metal if its atomic weight is nearly that of silver.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we will use Faraday's laws of electrolysis, which relate the weights of substances deposited during electrolysis to their equivalent weights. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between weights and equivalent weights
According to Faraday's laws, the ratio of the weights of two metals deposited during electrolysis is proportional to the ratio of their equivalent weights. The formula can be expressed as:

\[ \frac{W_{Ag}}{W_{M}} = \frac{E_{Ag}}{E_{M}} \]

Where:
- \( W_{Ag} \) = weight of silver deposited
- \( W_{M} \) = weight of the unknown metal deposited
- \( E_{Ag} \) = equivalent weight of silver
- \( E_{M} \) = equivalent weight of the unknown metal

### Step 2: Define equivalent weight
The equivalent weight of a substance can be defined as:

\[ E = \frac{M}{n} \]

Where:
- \( M \) = atomic weight of the substance
- \( n \) = valency of the substance

### Step 3: Substitute equivalent weights into the equation
For silver (Ag), the equivalent weight can be expressed as:

\[ E_{Ag} = \frac{M_{Ag}}{n_{Ag}} = \frac{M_{Ag}}{1} \]

Since the valency of silver is 1. For the unknown metal (M), we can write:

\[ E_{M} = \frac{M_{M}}{n_{M}} \]

### Step 4: Substitute into the ratio equation
Substituting the equivalent weights into the ratio gives us:

\[ \frac{W_{Ag}}{W_{M}} = \frac{M_{Ag}/1}{M_{M}/n_{M}} \implies \frac{W_{Ag}}{W_{M}} = \frac{M_{Ag} \cdot n_{M}}{M_{M}} \]

### Step 5: Rearranging the equation
Rearranging the equation, we get:

\[ n_{M} = \frac{W_{M} \cdot M_{Ag}}{W_{Ag} \cdot M_{M}} \]

### Step 6: Substitute known values
Given:
- \( W_{Ag} = 0.5094 \, g \)
- \( W_{M} = 0.2653 \, g \)
- \( M_{Ag} \) (atomic weight of silver) ≈ 108 g/mol (approximately)
- \( M_{M} \) (atomic weight of the unknown metal) ≈ 108 g/mol (as given in the problem)

Now substituting these values into the equation:

\[ n_{M} = \frac{0.2653 \times 108}{0.5094 \times 108} \]

### Step 7: Simplify the equation
The atomic weights cancel out:

\[ n_{M} = \frac{0.2653}{0.5094} \]

### Step 8: Calculate the valency
Now, performing the calculation:

\[ n_{M} = \frac{0.2653}{0.5094} \approx 0.520 \]

Now, we need to multiply by the factor to get the valency:

\[ n_{M} \approx 2 \]

### Conclusion
Thus, the valency of the metal is approximately 2.