A solution of sodium chloride discharge `6.0 xx 10^(16)Na^(+) and 4.5 xx 10^(16)Cl^(-1)` lons in `2` seconds. What is the current passing through the solution ?

Explanation

To find the current passing through the sodium chloride solution, we can follow these steps:

### Step 1: Calculate the total number of ions
We have:
- Sodium ions (Na⁺): \(6.0 \times 10^{16}\)
- Chloride ions (Cl⁻): \(4.5 \times 10^{16}\)

Total number of ions = \(6.0 \times 10^{16} + 4.5 \times 10^{16} = 10.5 \times 10^{16}\)

### Step 2: Calculate the total charge
The charge of one ion (either Na⁺ or Cl⁻) is equal to the charge of one electron, which is approximately \(1.6 \times 10^{-19} \text{ C}\).

Total charge \(Q\) can be calculated as:
\[ Q = \text{Total number of ions} \times \text{Charge of one ion} \]
\[ Q = (10.5 \times 10^{16}) \times (1.6 \times 10^{-19}) \text{ C} \]

### Step 3: Calculate the total charge
Now, let's compute the total charge:
\[ Q = 10.5 \times 1.6 \times 10^{16} \times 10^{-19} \]
\[ Q = 16.8 \times 10^{-3} \text{ C} \]

### Step 4: Calculate the current
Current \(I\) is defined as the charge per unit time:
\[ I = \frac{Q}{t} \]
where \(t = 2 \text{ seconds}\).

Substituting the values:
\[ I = \frac{16.8 \times 10^{-3}}{2} \text{ A} \]
\[ I = 8.4 \times 10^{-3} \text{ A} \]

### Step 5: Convert to milliAmperes
To convert amperes to milliAmperes, we multiply by \(1000\):
\[ I = 8.4 \times 10^{-3} \text{ A} = 8.4 \text{ mA} \]

### Final Answer
The current passing through the solution is \(8.4 \text{ mA}\).

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