The same quantity of electrical charge deposited `0.583g` of Ag when passed through `AgNO_(3),AuCl_(3)` solution calculate the weight of gold formed. (At weighht of `Au=197gmol)`

Explanation

To solve the problem, we will use the concept of equivalent weights and the relationship between the quantities of substances deposited during electrolysis.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Weight of silver (Ag) deposited: \( W_{Ag} = 0.583 \, \text{g} \)
- Atomic weight of gold (Au): \( W_{Au} = 197 \, \text{g/mol} \)
- Atomic weight of silver (Ag): \( W_{Ag} = 108 \, \text{g/mol} \)
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