`500mL` of `1N` solution of `CuCl_(2)` was electrolyszed with a current fo 2 amperes for 1 hour. What is the normality of the remaining `CuCl_(2)` solution ?
a 0.85
b 0.15
c 0.30
d 1.0

Options

A. `0.85`
B. `0.15`
C. `0.30`
D. `1.0`

Explanation

To solve the problem, we need to determine the normality of the remaining `CuCl₂` solution after electrolysis. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the initial milli-equivalents of `CuCl₂`
The formula to calculate milli-equivalents (mEq) is:
\[ \text{mEq} = \text{Volume (L)} \times \text{Normality (N)} \]
Given:
- Volume = 500 mL = 0.5 L
- Normality = 1 N

Calculating:
\[ \text{mEq} = 0.5 \, \text{L} \times 1 \, \text{N} = 0.5 \, \text{mEq} = 500 \, \text{mEq} \]

### Step 2: Calculate the total electricity passed during electrolysis
The total electricity (in coulombs) can be calculated using the formula:
\[ Q = I \times t \]
Where:
- \( I = 2 \, \text{A} \) (current)
- \( t = 1 \, \text{hour} = 3600 \, \text{seconds} \)

Calculating:
\[ Q = 2 \, \text{A} \times 3600 \, \text{s} = 7200 \, \text{C} \]

### Step 3: Calculate the number of equivalents of `CuCl₂` reacted
Using Faraday's law, the number of equivalents can be calculated as:
\[ \text{Equivalents} = \frac{Q}{F} \]
Where \( F \) (Faraday's constant) is approximately \( 96500 \, \text{C/mol} \).

Calculating:
\[ \text{Equivalents} = \frac{7200 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.0746 \, \text{equivalents} = 74.6 \, \text{mEq} \]

### Step 4: Calculate the remaining milli-equivalents of `CuCl₂`
Subtract the milli-equivalents reacted from the initial milli-equivalents:
\[ \text{Remaining mEq} = \text{Initial mEq} - \text{Reacted mEq} \]
Calculating:
\[ \text{Remaining mEq} = 500 \, \text{mEq} - 74.6 \, \text{mEq} = 425.4 \, \text{mEq} \]

### Step 5: Calculate the normality of the remaining solution
Normality can be calculated using the formula:
\[ \text{Normality} = \frac{\text{Remaining mEq}}{\text{Volume (L)}} \]
Given that the volume remains 0.5 L:
\[ \text{Normality} = \frac{425.4 \, \text{mEq}}{0.5 \, \text{L}} = 850.8 \, \text{mEq/L} = 0.85 \, \text{N} \]

### Final Answer:
The normality of the remaining `CuCl₂` solution is **0.85 N**.