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 of how much volume of hydrogen will be liberated by the same quantity of electricity that deposited 0.54 g of Ag from AgNO₃ solution, we can follow these steps:

### Step 1: Determine the number of equivalents of silver deposited
The equivalent mass of silver (Ag) can be calculated using its atomic mass. The atomic mass of silver is 108 g/mol, and since it gains one electron during the reduction process, its equivalent mass is:

\[ \text{Equivalent mass of Ag} = \frac{\text{Atomic mass of Ag}}{n} = \frac{108 \text{ g/mol}}{1} = 108 \text{ g/equiv} \]

Now, we can find the number of equivalents of silver deposited:

\[ \text{Number of equivalents of Ag} = \frac{\text{mass of Ag}}{\text{equivalent mass of Ag}} = \frac{0.54 \text{ g}}{108 \text{ g/equiv}} = 0.005 \text{ equiv} \]

### Step 2: Relate the equivalents of hydrogen to the equivalents of silver
From the electrochemical reaction, we know that 1 equivalent of silver corresponds to 1 equivalent of hydrogen (H₂). Therefore, the number of equivalents of hydrogen liberated will also be 0.005 equiv.

### Step 3: Calculate the moles of hydrogen
Since 1 equivalent of hydrogen corresponds to 1 mole of H₂, the moles of hydrogen liberated will also be:

\[ \text{Moles of H₂} = 0.005 \text{ mol} \]

### Step 4: Use the Ideal Gas Law to find the volume of hydrogen
The Ideal Gas Law is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin

First, convert the pressure from mmHg to atm:

\[ P = \frac{728 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.9579 \text{ atm} \]

Convert the temperature from Celsius to Kelvin:

\[ T = 27 + 273 = 300 \text{ K} \]

Now, substitute the values into the Ideal Gas Law:

\[ V = \frac{nRT}{P} = \frac{0.005 \text{ mol} \times 0.0821 \text{ L·atm/(K·mol)} \times 300 \text{ K}}{0.9579 \text{ atm}} \]

Calculating this gives:

\[ V = \frac{0.12315 \text{ L·atm}}{0.9579 \text{ atm}} \approx 0.1285 \text{ L} \]

### Step 5: Convert volume to milliliters
To convert liters to milliliters, multiply by 1000:

\[ V \approx 0.1285 \text{ L} \times 1000 \text{ mL/L} \approx 128.5 \text{ mL} \]

### Final Answer
The volume of hydrogen liberated is approximately **128.5 mL**.

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