The volume of `H_(2)` gas at NTP obtained by passing 4 amperes through acidified `H_(2)O` for 30 minutes is

Options

A. 0.836L
B. 0.0432L
C. 0.1672L
D. 5.6L

Explanation

To solve the problem of finding the volume of \( H_2 \) gas at NTP obtained by passing 4 amperes through acidified \( H_2O \) for 30 minutes, we can follow these steps:

### Step 1: Calculate the total charge (Q) passed through the solution
The charge can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes (A)
- \( t \) is the time in seconds (s)

Given:
- \( I = 4 \, \text{A} \)
- \( t = 30 \, \text{minutes} = 30 \times 60 \, \text{s} = 1800 \, \text{s} \)

Now, substituting the values:
\[ Q = 4 \, \text{A} \times 1800 \, \text{s} = 7200 \, \text{C} \]

### Step 2: Relate the charge to moles of \( H_2 \) produced
From electrolysis, we know that:
\[ 2 \, \text{moles of electrons} \rightarrow 1 \, \text{mole of } H_2 \]
The charge required to produce 1 mole of \( H_2 \) is given by Faraday's law:
\[ 1 \, \text{mole of electrons} = 96500 \, \text{C} \]
Thus, for 2 moles of electrons:
\[ 2 \, \text{moles of electrons} = 2 \times 96500 \, \text{C} = 193000 \, \text{C} \]

### Step 3: Calculate the moles of \( H_2 \) produced
Using the total charge (Q) calculated earlier:
\[ \text{Moles of } H_2 = \frac{Q}{\text{Charge required for 1 mole of } H_2} \]
\[ \text{Moles of } H_2 = \frac{7200 \, \text{C}}{193000 \, \text{C}} = 0.0373 \, \text{moles} \]

### Step 4: Calculate the volume of \( H_2 \) at NTP
At NTP (Normal Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the volume of \( H_2 \) produced can be calculated as:
\[ \text{Volume of } H_2 = \text{Moles of } H_2 \times 22.4 \, \text{L/mole} \]
\[ \text{Volume of } H_2 = 0.0373 \, \text{moles} \times 22.4 \, \text{L/mole} = 0.835 \, \text{L} \]

### Final Answer
The volume of \( H_2 \) gas at NTP obtained is \( 0.835 \, \text{L} \).

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