The number of atoms of `Ca` that will be deposited from a solution of `CaCl_(2)` by a current of `25mA` for `60s` will be

Options

A. `4.68xx10^(18)`
B. `4.68xx10^(15)`
C. `4.68xx10^(10)`
D. `2.34xx10^(15)`

Explanation

To solve the problem of how many atoms of calcium (Ca) will be deposited from a solution of calcium chloride (CaCl₂) by a current of 25 mA for 60 seconds, we can follow these steps:

### Step 1: Calculate the total charge (Q)
The total charge can be calculated using the formula:
\[ Q = I \times t \]
Where:
- \( I \) is the current in amperes (A)
- \( t \) is the time in seconds (s)

Given:
- \( I = 25 \, \text{mA} = 25 \times 10^{-3} \, \text{A} \)
- \( t = 60 \, \text{s} \)

Now substituting the values:
\[ Q = 25 \times 10^{-3} \, \text{A} \times 60 \, \text{s} \]
\[ Q = 1.5 \, \text{C} \]

### Step 2: Calculate the number of moles of electrons (n)
Using Faraday's law, the total charge can also be expressed as:
\[ Q = n \times F \]
Where:
- \( n \) is the number of moles of electrons
- \( F \) is Faraday's constant, approximately \( 96500 \, \text{C/mol} \)

Rearranging the formula to find \( n \):
\[ n = \frac{Q}{F} \]

Substituting the values:
\[ n = \frac{1.5 \, \text{C}}{96500 \, \text{C/mol}} \]
\[ n \approx 1.55 \times 10^{-5} \, \text{mol} \]

### Step 3: Relate moles of electrons to moles of calcium
The reaction for the deposition of calcium from calcium ions is:
\[ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \]
From this reaction, we see that 2 moles of electrons are required to deposit 1 mole of calcium.

Thus, the moles of calcium deposited can be calculated as:
\[ \text{Moles of Ca} = \frac{n}{2} = \frac{1.55 \times 10^{-5}}{2} \]
\[ \text{Moles of Ca} \approx 7.75 \times 10^{-6} \, \text{mol} \]

### Step 4: Calculate the number of atoms of calcium
To find the number of atoms, we can use Avogadro's number:
\[ N = \text{Moles of Ca} \times N_A \]
Where \( N_A \) (Avogadro's number) is approximately \( 6.022 \times 10^{23} \, \text{atoms/mol} \).

Substituting the values:
\[ N = 7.75 \times 10^{-6} \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \]
\[ N \approx 4.67 \times 10^{18} \, \text{atoms} \]

### Final Answer
The number of atoms of calcium deposited is approximately:
\[ \boxed{4.67 \times 10^{18}} \]

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