A solution contains `A^(+)` and `B^(+)` in such a concentration that both deposit simultaneously. If current of `9.65` amp was passed through `100ml` solution for 55 seconds then find the final concentration of `A^(+)` ion if initial concentration of `B^(+)` is `0.1M`.
[Fill your answer by multiplying it with `10^(3)]`.
Given: `{:(A^(+)+e^(-)rarrA,,E^(@) =- 0.5 "volt"),(B^(+)+e^(-)rarrB,,E^(@) =- 0.56 "volt"),((2.303RT)/(F) =0.06,,):}`

Explanation

To solve the problem step by step, we will follow the logical sequence based on the information provided in the question.

### Step 1: Understand the Given Information
We have a solution containing ions \( A^+ \) and \( B^+ \) with the following details:
- Current (I) = 9.65 A
- Volume of solution = 100 mL = 0.1 L
- Time (t) = 55 seconds
- Initial concentration of \( B^+ \) = 0.1 M
- Standard reduction potentials:
- \( E^\circ(A^+/A) = -0.5 \) V
- \( E^\circ(B^+/B) = -0.56 \) V
- \( \frac{2.303RT}{F} = 0.06 \)

### Step 2: Calculate the Total Charge (Q)
Using the formula for charge:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 9.65 \, \text{A} \times 55 \, \text{s} = 530.75 \, \text{C} \]

### Step 3: Calculate the Number of Moles of Electrons Transferred
Using Faraday's law, the number of moles of electrons (n) can be calculated as:
\[ n = \frac{Q}{F} \]
Where \( F = 96500 \, \text{C/mol} \):
\[ n = \frac{530.75}{96500} \approx 0.0055 \, \text{mol} \]

### Step 4: Determine the Equivalent of \( A^+ \)
Since the reaction involves the transfer of one electron (n-factor = 1), the equivalent of \( A^+ \) is equal to the moles of electrons transferred:
\[ \text{Equivalent of } A^+ = n = 0.0055 \, \text{mol} \]

### Step 5: Calculate the Final Concentration of \( A^+ \)
To find the concentration of \( A^+ \), we first need to convert the moles of \( A^+ \) into concentration:
\[ \text{Concentration (C)} = \frac{\text{moles}}{\text{volume (L)}} \]
The volume of the solution is 0.1 L:
\[ C_{A^+} = \frac{0.0055}{0.1} = 0.055 \, \text{M} \]

### Step 6: Convert to Required Format
The question asks for the final concentration of \( A^+ \) multiplied by \( 10^3 \):
\[ C_{A^+} \times 10^3 = 0.055 \times 10^3 = 55 \]

### Final Answer
The final concentration of \( A^+ \) ion, multiplied by \( 10^3 \), is:
\[ \boxed{55} \]