1 L aqueous solution of `H_(2)SO_(4)` contains 0.02 m mol `H_(2)SO_(4)` . 50 % of this solution is diluted with deionized water to give 1 L solution (A) . In solution (A) , 001 m mol of `H_(2)SO_(4)` are added . Total m moles of `H_(2)SO_(4)` in the final solution is _________ `xx 10^(3)` m moles

Explanation

To solve the problem step by step, we will follow the instructions provided in the question.

### Step 1: Calculate the initial amount of `H2SO4` in the solution
The initial solution contains 0.02 millimoles of `H2SO4` in 1 L of aqueous solution.

### Step 2: Determine the amount of `H2SO4` after dilution
Since 50% of the solution is diluted with deionized water, we need to find out how much `H2SO4` remains in the diluted solution.

- 50% of the original solution is taken, which means:
\[ \text{Amount of } H2SO4 \text{ in 0.5 L} = \frac{0.02 \text{ m mol}}{2} = 0.01 \text{ m mol} \]

Now, this 0.01 m mol of `H2SO4` is diluted to make a total volume of 1 L (solution A).

### Step 3: Add additional `H2SO4` to solution A
In solution A, we add an additional 0.001 m mol of `H2SO4`.

### Step 4: Calculate the total amount of `H2SO4` in the final solution
Now, we need to sum the amount of `H2SO4` in solution A and the amount added:
\[ \text{Total } H2SO4 = 0.01 \text{ m mol} + 0.001 \text{ m mol} = 0.011 \text{ m mol} \]

### Step 5: Convert to the required format
The final answer needs to be expressed in the form of `xx × 10^3` m moles.

To express 0.011 m mol in this format:
\[ 0.011 \text{ m mol} = 11 \times 10^{-3} \text{ m mol} \]

Thus, the total m moles of `H2SO4` in the final solution is:
\[ \text{Final answer} = 11 \]

### Final Answer
The total m moles of `H2SO4` in the final solution is **11 × 10^3 m moles**.

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