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 solve the problem, we will use Faraday's first law of electrolysis, which states that the amount of substance deposited at an electrode is directly proportional to the quantity of electric charge passed through the electrolyte.

### Step-by-Step Solution:

1. **Calculate the Total Charge (Q)**:
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where \(I\) is the current in amperes and \(t\) is the time in seconds.

Given:
- Current \(I = 1 \, \text{A}\)
- Time \(t = 16 \, \text{min} \, 5 \, \text{sec} = (16 \times 60) + 5 = 965 \, \text{sec}\)

Now, substituting the values:
\[ Q = 1 \, \text{A} \times 965 \, \text{sec} = 965 \, \text{C} \]

2. **Calculate the Number of Equivalents of Copper Deposited**:
Using Faraday's first law:
\[ \frac{W}{E} = \frac{Q}{96500} \]
where \(W\) is the weight of copper deposited and \(E\) is the equivalent weight of copper.

The equivalent weight of copper (Cu) is given by:
\[ E = \frac{M}{n} \]
where \(M\) is the molar mass of copper (approximately 63.5 g/mol) and \(n\) is the number of electrons transferred per atom of copper (which is 2 for Cu²⁺).

Thus:
\[ E = \frac{63.5}{2} = 31.75 \, \text{g/equiv} \]

3. **Substituting Values to Find W**:
Now substituting the values into the equation:
\[ \frac{W}{31.75} = \frac{965}{96500} \]
Rearranging gives:
\[ W = 31.75 \times \frac{965}{96500} \]
Simplifying:
\[ W = 31.75 \times 0.01 = 0.3175 \, \text{g} \]

4. **Calculate the Normality of CuCl₂ Solution**:
Normality (N) is defined as the number of equivalents of solute per liter of solution. Since we have calculated the weight of copper deposited, we can find the number of equivalents:
\[ \text{Number of equivalents} = \frac{W}{E} = \frac{0.3175}{31.75} = 0.01 \, \text{equiv} \]

Since the volume of the solution is 1 liter, the normality (N) is:
\[ N = \text{Number of equivalents} = 0.01 \, \text{N} \]

Thus, the normality of the CuCl₂ solution is **0.01 N**.