On passing electric current of one ampere for 16 min and 5 sec through one litre solution of `CuCl_(2)` , all copper of solution was deposited at cathode. The normality of `CuCl_(2)` solution was:

Options

A. `0.2` N
B. `0.01` N
C. `0.01` N
D. `0.02` N

Answer

To find the normality of the `CuCl₂` solution after passing an electric current, we will follow these steps:

### Step 1: Convert Time to Seconds
First, we need to convert the time given in minutes and seconds to seconds.
- Time = 16 minutes and 5 seconds
- Convert minutes to seconds:
\[ 16 \text{ minutes} = 16 \times 60 = 960 \text{ seconds} \]
- Total time in seconds:
\[ 960 \text{ seconds} + 5 \text{ seconds} = 965 \text{ seconds} \]

### Step 2: Use Faraday's Law to Calculate Weight of Copper Deposited
According to Faraday's law, the weight of the substance deposited is given by:
\[ \text{Weight} = \frac{I \cdot t}{n \cdot F} \]
Where:
- \( I \) = current in amperes = 1 A
- \( t \) = time in seconds = 965 s
- \( n \) = number of electrons transferred per ion (for Cu²⁺, n = 2)
- \( F \) = Faraday's constant = 96500 C/mol

Substituting the values:
\[ \text{Weight} = \frac{1 \cdot 965}{2 \cdot 96500} \]
Calculating the weight:
\[ \text{Weight} = \frac{965}{193000} \approx 0.005 \text{ kg} = 0.3175 \text{ grams} \]

### Step 3: Calculate Number of Moles of Copper
To find the number of moles of copper deposited, we use the formula:
\[ \text{Number of moles} = \frac{\text{Weight}}{\text{Molar mass}} \]
Where the molar mass of copper (Cu) is 63.5 g/mol.
\[ \text{Number of moles} = \frac{0.3175 \text{ g}}{63.5 \text{ g/mol}} \approx 0.005 \text{ moles} \]

### Step 4: Calculate Molarity of the Solution
Molarity (M) is calculated using the formula:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume of solution in liters}} \]
Given that the volume of the solution is 1 liter:
\[ \text{Molarity} = \frac{0.005 \text{ moles}}{1 \text{ L}} = 0.005 \text{ M} \]

### Step 5: Calculate Normality of the Solution
Normality (N) is related to molarity by the equation:
\[ \text{Normality} = \text{Molarity} \times n \]
Where \( n \) is the number of equivalents (valency) of the solute. For Cu²⁺, \( n = 2 \):
\[ \text{Normality} = 0.005 \text{ M} \times 2 = 0.01 \text{ N} \]

### Conclusion
The normality of the `CuCl₂` solution is:
\[ \text{Normality} = 0.01 \text{ N} \]