To solve the problem, we need to follow these steps:
### Step 1: Determine the equivalent mass of silver (Ag)
The equivalent mass of silver (Ag) can be calculated using the formula:
\[
\text{Equivalent mass of Ag} = \frac{\text{Molar mass of Ag}}{n}
\]
where \( n \) is the number of electrons transferred per atom of silver. For silver, \( n = 1 \) (since Ag^+ gains one electron to become Ag). The molar mass of silver is approximately 108 g/mol.
\[
\text{Equivalent mass of Ag} = \frac{108 \, \text{g/mol}}{1} = 108 \, \text{g/equiv}
\]
### Step 2: Calculate the number of equivalents of Ag deposited
Using the mass of silver deposited (0.54 g), we can find the number of equivalents of silver:
\[
\text{Number of equivalents of Ag} = \frac{\text{mass of Ag}}{\text{Equivalent mass of Ag}} = \frac{0.54 \, \text{g}}{108 \, \text{g/equiv}} = 0.005 \, \text{equiv}
\]
### Step 3: Determine the equivalent mass of hydrogen (H2)
The equivalent mass of hydrogen (H2) is calculated as follows:
\[
\text{Equivalent mass of H2} = \frac{\text{Molar mass of H2}}{n}
\]
For hydrogen, \( n = 2 \) (since H2 releases 2 electrons). The molar mass of H2 is approximately 2 g/mol.
\[
\text{Equivalent mass of H2} = \frac{2 \, \text{g/mol}}{2} = 1 \, \text{g/equiv}
\]
### Step 4: Calculate the mass of hydrogen liberated
Using the number of equivalents of silver, we can find the mass of hydrogen liberated:
\[
\text{Mass of H2} = \text{Number of equivalents of H2} \times \text{Equivalent mass of H2}
\]
Since the number of equivalents of H2 is the same as that of Ag:
\[
\text{Mass of H2} = 0.005 \, \text{equiv} \times 1 \, \text{g/equiv} = 0.005 \, \text{g}
\]
### Step 5: Calculate the volume of hydrogen at given conditions
Using the ideal gas law:
\[
PV = nRT
\]
We need to convert the mass of hydrogen to moles:
\[
\text{Moles of H2} = \frac{\text{mass of H2}}{\text{molar mass of H2}} = \frac{0.005 \, \text{g}}{2 \, \text{g/mol}} = 0.0025 \, \text{mol}
\]
Now, we can calculate the volume (V) using the conditions given:
- Pressure (P) = 750 mmHg = \( \frac{750}{760} \, \text{atm} \approx 0.9868 \, \text{atm} \)
- Temperature (T) = 27°C = 300 K
- R (ideal gas constant) = 0.0821 L·atm/(K·mol)
Now substituting into the ideal gas equation:
\[
V = \frac{nRT}{P} = \frac{0.0025 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}}{0.9868 \, \text{atm}}
\]
Calculating the volume:
\[
V \approx \frac{0.061575}{0.9868} \approx 0.06234 \, \text{L} = 62.34 \, \text{mL}
\]
### Final Answer
The volume of hydrogen liberated is approximately **62.34 mL**.
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