To solve the problem, we will follow these steps:
### Step 1: Calculate the total charge passed through the solution.
The formula to calculate the total charge (Q) is given by:
\[ Q = I \times t \]
Where:
- \( I \) is the current in amperes (A)
- \( t \) is the time in seconds (s)
Given:
- \( I = 96.5 \, \text{A} \)
- \( t = 18 \, \text{minutes} = 18 \times 60 \, \text{s} = 1080 \, \text{s} \)
Now, substituting the values:
\[ Q = 96.5 \, \text{A} \times 1080 \, \text{s} = 104220 \, \text{C} \]
### Step 2: Determine the charge required to deposit one mole of nickel.
From Faraday's laws of electrolysis, we know that 1 mole of nickel (Ni) requires 2 moles of electrons for reduction. The charge required for 1 mole of nickel is:
\[ Q_{\text{Ni}} = n \times F \]
Where:
- \( n = 2 \) (moles of electrons)
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
Calculating the charge:
\[ Q_{\text{Ni}} = 2 \times 96500 = 193000 \, \text{C} \]
### Step 3: Calculate the number of moles of nickel deposited.
Using the total charge calculated in Step 1, we can find the number of moles of nickel deposited (n):
\[ n = \frac{Q}{Q_{\text{Ni}}} \]
Substituting the values:
\[ n = \frac{104220 \, \text{C}}{193000 \, \text{C/mol}} \approx 0.54 \, \text{mol} \]
### Step 4: Calculate the initial moles of nickel in the solution.
The initial concentration of nickel nitrate (\( Ni(NO_3)_2 \)) is given as 2 M in 500 mL of solution. To find the initial moles of nickel:
\[ \text{Initial moles} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Initial moles} = 2 \, \text{mol/L} \times 0.5 \, \text{L} = 1 \, \text{mol} \]
### Step 5: Calculate the remaining moles of nickel after electrolysis.
The remaining moles of nickel in the solution after electrolysis can be calculated as:
\[ \text{Remaining moles} = \text{Initial moles} - \text{Deposited moles} \]
\[ \text{Remaining moles} = 1 \, \text{mol} - 0.54 \, \text{mol} = 0.46 \, \text{mol} \]
### Step 6: Calculate the molarity of the solution after electrolysis.
Finally, we can find the molarity of the solution after electrolysis:
\[ \text{Molarity} = \frac{\text{Remaining moles}}{\text{Volume (L)}} \]
\[ \text{Molarity} = \frac{0.46 \, \text{mol}}{0.5 \, \text{L}} = 0.92 \, \text{M} \]
### Final Answer:
The molarity of the solution after electrolysis is **0.92 M**.
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