When a quantity of electricity is passed through `CuSO_(4)` solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of `H_(2)` liberated at STP will be : (given atomic weight of Cu=64)

Options

A. `4.0cm^(3)`
B. `224cm^(3)`
C. `604cm^(3)`
D. =`(8.0cm^3)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of copper deposited.
Given:
- Mass of copper deposited (m) = 0.16 g
- Atomic weight of copper (Cu) = 64 g/mol

Using the formula for number of moles:
\[ \text{Number of moles of Cu} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.16 \, \text{g}}{64 \, \text{g/mol}} = 0.0025 \, \text{mol} \]

### Step 2: Determine the amount of electricity (in Faraday) used to deposit the copper.
From electrochemistry, we know that:
- 1 mole of Cu requires 2 Faraday to deposit.

Thus, for 0.0025 moles of Cu:
\[ \text{Electricity (in Faraday)} = 0.0025 \, \text{mol} \times 2 \, \text{Faraday/mol} = 0.005 \, \text{Faraday} \]

### Step 3: Calculate the number of moles of hydrogen gas (H₂) produced from the same quantity of electricity.
The reaction for hydrogen gas production is:
\[ 2H^+ + 2e^- \rightarrow H_2 \]
This means that 1 mole of H₂ is produced from 2 Faraday.

Let \( y \) be the number of moles of H₂ produced. Therefore:
\[ 2y = 0.005 \, \text{Faraday} \]
Solving for \( y \):
\[ y = \frac{0.005}{2} = 0.0025 \, \text{mol} \]

### Step 4: Calculate the volume of hydrogen gas liberated at STP.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters (or 22400 mL).

Using the number of moles of H₂ calculated:
\[ \text{Volume of } H_2 = \text{Number of moles} \times \text{Molar volume at STP} \]
\[ \text{Volume of } H_2 = 0.0025 \, \text{mol} \times 22400 \, \text{mL/mol} = 56 \, \text{mL} \]

### Final Answer:
The volume of hydrogen gas liberated at STP is **56 mL** (or **56 cm³**).

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