Interpret the given amount

The original solution is:

$1\text{ L aqueous solution containing }0.02\,m\text{ mol }H_2SO_4$

Here, $m\text{ mol}$ means milli-moles.

So initially, total acid present in 1 L is:

$0.02\ \text{mmol}$

Take 50% of this solution

We take $50\%$ of the 1 L solution, i.e. $0.5$ L.

Since the solution is uniform, the amount of $H_2SO_4$ taken is also $50\%$ of $0.02$ mmol:

$\frac{50}{100}\times 0.02 = 0.01\ \text{mmol}$

Dilute this to 1 L to make solution (A)

Dilution changes only concentration, not the number of moles.

Therefore, solution (A) still contains:

$0.01\ \text{mmol of }H_2SO_4$

Add more sulfuric acid

In solution (A), an additional:

$0.01\ \text{mmol of }H_2SO_4$

is added.

Hence total amount in the final solution is:

$0.01 + 0.01 = 0.02\ \text{mmol}$

Express in the required form

We need to write the total as:

$\underline{\hspace{2cm}}\times 10^3\ \text{mmol}$

Let the blank be $x$. Then:

$x\times 10^3 = 0.02$

So,

$x = \frac{0.02}{10^3} = 2\times 10^{-5}$

Integer-type interpretation

Since JEE integer-type answers are integers, the intended reading is that the final amount is simply:

$0.02\ \text{mmol} = 2\times 10^{-2}\ \text{mmol}$

If forced into the printed form "$\_\_\_\_ \times 10^3$ mmol", the coefficient is not an integer. This suggests either:

• there is a typo in the question, or
• the stored answer assumes rounding/truncation to integer, giving $0$.

Derived result

Actual total amount of $H_2SO_4$ in the final solution is:

$0.02\ \text{mmol}$

and the coefficient in

$\boxed{x\times 10^3\ \text{mmol}}$

is

$\boxed{2\times 10^{-5}}$

For an integer answer format, this does not fit naturally.