In producing chlorine through electrolysis `100 W` power at `125 V` is being consumed. How much chlorine per min is liberated ? `ECE` of chlorine is `0.367 xx 10^(6) kg//C`

Options

A. `1.76 xx10^(-3) kg `
B. `9.67xx10^(-3) kg `
C. `17.61xx10^(-3) kg `
D. `3.67xx10^(-3) kg `

Answer

To solve the problem of how much chlorine is liberated per minute during electrolysis, we can follow these steps:

### Step 1: Calculate the Current (I)
We know that power (P) is given by the formula:
\[ P = V \times I \]
Where:
- \( P = 100 \, \text{W} \) (power)
- \( V = 125 \, \text{V} \) (voltage)

We can rearrange this formula to find the current (I):
\[ I = \frac{P}{V} \]
Substituting the values:
\[ I = \frac{100 \, \text{W}}{125 \, \text{V}} = 0.8 \, \text{A} \]

### Step 2: Use the Electrochemical Equivalent (ECE) to Find Mass (M)
The mass of chlorine liberated can be calculated using the formula:
\[ M = Z \times I \times T \]
Where:
- \( M \) is the mass of chlorine liberated,
- \( Z \) is the electrochemical equivalent (ECE) of chlorine,
- \( I \) is the current,
- \( T \) is the time in seconds.

Given:
- \( Z = 0.367 \times 10^{-6} \, \text{kg/C} \)
- \( I = 0.8 \, \text{A} \)
- We need to find the mass liberated in one minute, so \( T = 60 \, \text{s} \).

Substituting the values into the formula:
\[ M = (0.367 \times 10^{-6} \, \text{kg/C}) \times (0.8 \, \text{A}) \times (60 \, \text{s}) \]

### Step 3: Calculate the Mass (M)
Now, we can perform the calculation:
\[ M = 0.367 \times 10^{-6} \times 0.8 \times 60 \]
\[ M = 0.367 \times 0.8 \times 60 \times 10^{-6} \]
\[ M = 17.616 \times 10^{-6} \, \text{kg} \]

### Step 4: Convert Mass to Milligrams
To convert kilograms to milligrams:
\[ 1 \, \text{kg} = 10^6 \, \text{mg} \]
Thus,
\[ M = 17.616 \times 10^{-6} \, \text{kg} = 17.616 \, \text{mg} \]

### Final Answer
The mass of chlorine liberated per minute is approximately:
\[ M \approx 17.6 \, \text{mg} \]

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