A certain electricity deposited `0.54g` of Ag from `AgNO_(3)` solution what volume of hydrogen will the same quantity of electricity liberate at `27^(@)C` and `728mmMg` pressure?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
- Mass of silver deposited (Ag) = 0.54 g
- Temperature (T) = 27°C = 27 + 273 = 300 K
- Pressure (P) = 728 mm Hg

### Step 2: Determine the Equivalent Mass
The equivalent mass of silver (Ag) is 108 g/mol. The equivalent mass of hydrogen (H) is 1 g/mol.

### Step 3: Set Up the Mass Ratio
Using the relationship between the mass of hydrogen and the mass of silver deposited:
\[ \frac{\text{Mass of H}}{\text{Mass of Ag}} = \frac{\text{Equivalent mass of H}}{\text{Equivalent mass of Ag}} \]

Substituting the known values:
\[ \frac{\text{Mass of H}}{0.54 \, \text{g}} = \frac{1}{108} \]

### Step 4: Calculate the Mass of Hydrogen
Rearranging the equation gives:
\[ \text{Mass of H} = 0.54 \times \frac{1}{108} \]
Calculating this:
\[ \text{Mass of H} = \frac{0.54}{108} = 0.005 \, \text{g} \, \text{(or 5.0 x 10}^{-3} \text{g)} \]

### Step 5: Use the Ideal Gas Law to Find Volume
Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( n = \frac{\text{mass}}{\text{molar mass}} \)
- Molar mass of hydrogen (H₂) = 2 g/mol

Calculating the number of moles of hydrogen:
\[ n = \frac{0.005 \, \text{g}}{2 \, \text{g/mol}} = 0.0025 \, \text{mol} \]

### Step 6: Convert Pressure to Atmospheres
Convert pressure from mm Hg to atm:
\[ P = \frac{728 \, \text{mm Hg}}{760} = 0.9579 \, \text{atm} \]

### Step 7: Substitute Values into Ideal Gas Law
Now substituting values into the ideal gas law:
\[ V = \frac{nRT}{P} \]
Where \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \):
\[ V = \frac{(0.0025 \, \text{mol}) \times (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K})}{0.9579 \, \text{atm}} \]

### Step 8: Calculate the Volume
Calculating:
\[ V = \frac{(0.0025) \times (0.0821) \times (300)}{0.9579} \approx \frac{0.062675}{0.9579} \approx 0.0654 \, \text{L} \]

### Final Answer
The volume of hydrogen liberated is approximately **0.0654 liters**.

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