A current of `96.5` A is passed for 18 min between nickel electrodes in 500 mL solution of `2M Ni(NO_(3) ) _(2)` . The molarity of solution after electrolysis would be:

Options

A. `0.46` M
B. 0.92 M
C. `0.625` M
D. `1.25` M

Answer

To find the molarity of the nickel nitrate solution after electrolysis, we will follow these steps:

### Step 1: Calculate the initial moles of Ni(NO₃)₂
Given:
- Molarity (C) = 2 M
- Volume (V) = 500 mL = 0.5 L (since 1000 mL = 1 L)

Using the formula for moles:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles} = 2 \, \text{mol/L} \times 0.5 \, \text{L} = 1 \, \text{mol} \]

### Step 2: Calculate the total charge passed through the solution
Given:
- Current (I) = 96.5 A
- Time (t) = 18 min = 18 × 60 = 1080 seconds

Using the formula for charge:
\[ Q = I \times t \]
\[ Q = 96.5 \, \text{A} \times 1080 \, \text{s} = 104220 \, \text{C} \]

### Step 3: Determine how many moles of nickel are deposited
The reaction at the nickel electrode can be represented as:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni (s)} \]
From the reaction, we see that 1 mole of Ni requires 2 moles of electrons (2 Faraday).

Using Faraday's constant:
\[ F = 96500 \, \text{C/mol} \]
Thus, 2 moles of electrons correspond to:
\[ Q_{\text{required}} = 2 \times 96500 \, \text{C} = 193000 \, \text{C} \]

Now, we can find the moles of nickel deposited using the charge passed:
\[ \text{Moles of Ni deposited} = \frac{Q}{2F} = \frac{104220 \, \text{C}}{193000 \, \text{C/mol}} \approx 0.54 \, \text{mol} \]

### Step 4: Calculate the remaining moles of Ni(NO₃)₂ in solution
Initial moles of Ni(NO₃)₂ = 1 mol
Moles of Ni deposited = 0.54 mol

Remaining moles of Ni(NO₃)₂:
\[ \text{Remaining moles} = 1 - 0.54 = 0.46 \, \text{mol} \]

### Step 5: Calculate the final molarity of the solution
Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{Moles}}{\text{Volume in L}} \]
Volume = 0.5 L (500 mL converted to liters)

\[ \text{Final Molarity} = \frac{0.46 \, \text{mol}}{0.5 \, \text{L}} = 0.92 \, \text{mol/L} \]

### Final Answer
The molarity of the solution after electrolysis is **0.92 M**.

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