10 mL of 2 M NaOH solution is added to 20 mL of 1 M HCl solution kept in a beaker. Now, 10 mL of this mixture is poured into a volumetric flask of 100 mL containing 2 moles of HCl and made the volume upto the mark with distilled water. The solution in this flask is:

The relation between moles, molarity and volume is $$n = M \times V$$, where $$n$$ is in moles, $$M$$ in mol L$$^{-1}$$ and $$V$$ in litres.

Step 1 • Moles present before mixing
NaOH: $$M = 2\,\text{M},\; V = 10\,\text{mL}=0.01\,\text{L}$$
$$n_{NaOH} = 2 \times 0.01 = 0.02\,\text{mol}$$

HCl: $$M = 1\,\text{M},\; V = 20\,\text{mL}=0.02\,\text{L}$$
$$n_{HCl} = 1 \times 0.02 = 0.02\,\text{mol}$$

Step 2 • Neutralisation reaction
$$NaOH + HCl \rightarrow NaCl + H_2O$$
Since $$n_{NaOH}=n_{HCl}=0.02\,\text{mol}$$, both are completely consumed and we obtain $$0.02\,\text{mol}$$ NaCl.

Step 3 • Concentration of NaCl in the neutral mixture
Total volume after reaction $$=10\,\text{mL}+20\,\text{mL}=30\,\text{mL}=0.03\,\text{L}$$.
$$M_{NaCl} = \frac{0.02}{0.03} \approx 0.667\,\text{M}$$

Step 4 • Portion taken to the volumetric flask
Volume taken $$=10\,\text{mL}=0.01\,\text{L}$$.
Moles of NaCl transferred $$=0.667 \times 0.01 \approx 6.67\times10^{-3}\,\text{mol}$$.

Step 5 • HCl already present in the flask
The flask initially contains $$2\,\text{mol}$$ of HCl.

Step 6 • Final dilution to 100 mL
Final volume $$=100\,\text{mL}=0.1\,\text{L}$$.

Molarity of HCl:
$$M_{HCl} = \frac{2}{0.1} = 20\,\text{M}$$

Molarity of NaCl:
$$M_{NaCl} = \frac{6.67\times10^{-3}}{0.1} \approx 0.067\,\text{M}$$

Step 7 • Nature of the solution
The solution is a strongly acidic solution of $$20\,\text{M}$$ HCl (with a small amount of NaCl that does not affect acidity).

Hence the correct description is Option B - 20 M HCl solution.