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JEE Main 2025 Apr 3 Shift 2, Chemistry Q2: Stoichiometry

JEE Main2025Apr 3, Shift 2Chemistry
Q.

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 :

  1. A

    0.2 M NaCl solution

  2. B

    20 M HCl solution

  3. C

    10 M HCl solution

  4. D

    Neutral solution

Solution

First mixing step: moles of NaOH mol; moles of HCl mol. These neutralise exactly, leaving only NaCl in 30 mL of water. So the 10 mL aliquot transferred is essentially salt water and contributes no acid.

Second step: the 100 mL flask already contains 2 mol of HCl. After the aliquot is added and the volume is made up to 100 mL with distilled water, the total acid moles remain 2.

M.

The resulting solution is 20 M HCl.

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