A current strength of 0.965 amperes is passed through excess fused `AlCl_(3)` for 5 hours. How many litres of chlorine will be liberated at STP ? `(F=96500C)`

Options

A. 2.016
B. 1.008
C. 11.2
D. 20.16

Answer

To solve the problem of how many liters of chlorine will be liberated at STP when a current of 0.965 amperes is passed through excess fused AlCl₃ for 5 hours, we will follow these steps:

### Step 1: Calculate the total charge (Q) passed through the solution.
The formula to calculate charge (Q) is:
\[ Q = I \times t \]
where:
- \( I \) = current in amperes (0.965 A)
- \( t \) = time in seconds

First, convert the time from hours to seconds:
\[ t = 5 \text{ hours} \times 3600 \text{ seconds/hour} = 18000 \text{ seconds} \]

Now, calculate the charge:
\[ Q = 0.965 \, \text{A} \times 18000 \, \text{s} = 17370 \, \text{C} \]

### Step 2: Convert the charge to Faradays.
Using Faraday's constant \( F = 96500 \, \text{C} \):
\[ \text{Number of Faradays} = \frac{Q}{F} = \frac{17370 \, \text{C}}{96500 \, \text{C/F}} \approx 0.180 \, \text{F} \]

### Step 3: Determine the moles of chlorine produced.
From the electrolysis of AlCl₃, we know that:
- 1 mole of Cl₂ is produced by 2 Faradays of charge.

Thus, the moles of Cl₂ produced can be calculated as:
\[ \text{Moles of Cl₂} = \frac{\text{Number of Faradays}}{2} = \frac{0.180 \, \text{F}}{2} = 0.090 \, \text{moles} \]

### Step 4: Calculate the volume of chlorine gas at STP.
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the volume of chlorine gas produced is:
\[ \text{Volume of Cl₂} = \text{Moles of Cl₂} \times 22.4 \, \text{L/mole} \]
\[ \text{Volume of Cl₂} = 0.090 \, \text{moles} \times 22.4 \, \text{L/mole} \approx 2.016 \, \text{L} \]

### Final Answer:
The volume of chlorine gas liberated at STP is approximately **2.016 liters**.

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