JEE Main 2025 — 3 April, Evening Shift — Question 14

10 mL of 2 M NaOH reacts with 20 mL of 1 M HCl to give completely neutralised 30 mL of $\mathrm{NaCl}(20 \mathrm{~m}$ mole)

solution. 10 mL of this solution has $\frac{20}{30} \times 10=\frac{20}{3} \mathrm{~m} \mathrm{~mol}$ of NaCl

In Final 100 mL solution :

$$\begin{aligned} & {[\mathrm{HCl}]=\frac{2}{100} \times 1000=20 \mathrm{M}} \\& {[\mathrm{NaCl}]=\frac{20}{3 \times 100}=\frac{1}{15} \mathrm{M}} \end{aligned}$$