JEE Chemistry 2025 Question with Solution

Given:

• $10 \, \text{mL}$ of $2 \, \text{M}$ NaOH
• $20 \, \text{mL}$ of $1 \, \text{M}$ HCl
• $10 \, \text{mL}$ of the resulting mixture is transferred to a $100 \, \text{mL}$ volumetric flask containing $2$ moles of HCl

Find: The nature of the final solution in the flask.

First, determine the composition of the initial mixture.

The reaction is

Since NaOH and HCl are present in equal moles, they completely neutralize each other. So, the beaker contains a neutral NaCl solution with no excess acid or base.

Now, $10 \, \text{mL}$ of this neutral mixture is added to the volumetric flask already containing $2$ moles of HCl. This transferred portion does not contribute any extra HCl or NaOH.

The final volume is $100 \, \text{mL} = 0.1 \, \text{L}$. Therefore,

Therefore, the solution in the flask is $20 \, \text{M}$ HCl solution. The correct option is B.

Given: Neutralization between NaOH and HCl occurs first, and then part of that mixture is added to a flask containing HCl.

Find: Final concentration and nature of solution in the flask.

Step 1: In the beaker,

Step 2: Because the reaction ratio is $1:1$, both are consumed completely. Thus, the resulting solution contains only NaCl and water, so it is neutral.

Step 3: Taking $10 \, \text{mL}$ from this neutral solution means only neutral NaCl solution is transferred. The flask already has $2$ moles of HCl, so the acid present in the flask remains $2$ moles.

Step 4: After dilution to $100 \, \text{mL}$,

Hence, the final solution is $20 \, \text{M}$ HCl solution, so the correct option is B.