A current 0.5 ampere when passed through `AgNO_(3)` solution for 193 sec. Deposited 0.108 g of Ag. Find the equivalent weight of Ag.

Answer

To find the equivalent weight of silver (Ag) deposited from the electrolysis of silver nitrate (AgNO₃) solution, we can use the formula derived from Faraday's laws of electrolysis:

\[ \text{Weight deposited} = \frac{I \times t \times \text{Equivalent weight}}{F} \]

Where:
- \( I \) = current in amperes (A)
- \( t \) = time in seconds (s)
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- Weight deposited = mass of silver deposited (g)

Given:
- \( I = 0.5 \, A \)
- \( t = 193 \, s \)
- Weight deposited = \( 0.108 \, g \)

We need to rearrange the formula to solve for the equivalent weight of silver (Ag):

\[ \text{Equivalent weight} = \frac{\text{Weight deposited} \times F}{I \times t} \]

Now, substituting the known values into the equation:

1. Substitute the values into the equation:
\[ \text{Equivalent weight} = \frac{0.108 \, g \times 96500 \, C/mol}{0.5 \, A \times 193 \, s} \]

2. Calculate the denominator:
\[ 0.5 \, A \times 193 \, s = 96.5 \, C \]

3. Now, substitute this value back into the equation:
\[ \text{Equivalent weight} = \frac{0.108 \, g \times 96500 \, C/mol}{96.5 \, C} \]

4. Calculate the numerator:
\[ 0.108 \, g \times 96500 \, C/mol = 10429.4 \, g \cdot C/mol \]

5. Now, divide the numerator by the denominator:
\[ \text{Equivalent weight} = \frac{10429.4 \, g \cdot C/mol}{96.5 \, C} \approx 108 \, g/mol \]

Thus, the equivalent weight of silver (Ag) is approximately **108 g/mol**.