JEE MAIN - Chemistry (2025 - 3rd April Evening Shift - No. 7)
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 :
0.2 M NaCl solution
20 M HCl solution
Neutral solution
10 M HCl solution
Explanation
When $10 \mathrm{ml}, 2 \mathrm{M} \mathrm{NaOH}$ solution is added to 20 ml of 1 M HCl solution :
$\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}$
$$\begin{array}{rlrl} \text { Initial : MV } & =2 \times 0.1 & & M V=1 \times 0.2 \\ & =0.2 \text { mole } & & =0.2 \mathrm{~mole} \\ \text { Final } & 0 & 0 \end{array}$$
$\therefore$ Resulting solution becomes neutral.
Now when 10 mol of above solution is poured into a flask containing 2 mole HCl and made solution 100 ml will distilled water.
Molarity of $\mathrm{HCl}=\frac{2}{100} \times 1000=20$
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