1. First mixing in the beaker

Given:

• $10\,\text{mL}$ of $2\,\text{M}$ NaOH
• $20\,\text{mL}$ of $1\,\text{M}$ HCl

Calculate moles:

$n(\text{NaOH}) = 2 \times \frac{10}{1000} = 0.02\,\text{mol}$

$n(\text{HCl}) = 1 \times \frac{20}{1000} = 0.02\,\text{mol}$

Reaction:

$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Since moles are equal, both react completely.

So in the beaker, after reaction:

• NaOH left = $0$
• HCl left = $0$
• NaCl formed = $0.02\,\text{mol}$

Total volume in beaker:

$10 + 20 = 30\,\text{mL}$

Thus, the beaker contains $30\,\text{mL}$ of NaCl solution having $0.02\,\text{mol}$ NaCl.

2. Taking 10 mL of this mixture

From $30\,\text{mL}$ total solution, $10\,\text{mL}$ is taken, i.e. one-third of the total.

So moles of NaCl transferred:

$n(\text{NaCl transferred}) = 0.02 \times \frac{10}{30} = \frac{0.02}{3} \approx 0.00667\,\text{mol}$

3. Now this 10 mL is added to a 100 mL volumetric flask containing 2 moles of HCl

The flask already contains:

$n(\text{HCl}) = 2\,\text{mol}$

The transferred solution contains only NaCl and water; there is no NaOH left to neutralize HCl.

Then volume is made up to $100\,\text{mL} = 0.1\,\text{L}$.

So final molarity of HCl is:

$M = \frac{2}{0.1} = 20\,\text{M}$

NaCl is also present, but the dominant description among the options is 20 M HCl solution.

4. Checking options

• A: $0.2\,\text{M}$ NaCl solution
  Incorrect, because HCl is present in huge amount.

• B: $20\,\text{M}$ HCl solution
  Correct.

• C: Neutral solution
  Incorrect, strong acid is present.

• D: $10\,\text{M}$ HCl solution
  Incorrect, actual molarity is $20\,\text{M}$.

5. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

They agree.