JEE Main 2026 — Ionic Equilibrium Question with Solution
JEE Main 2026 (23 January Shift 1)
Question
of pure HCl was used to make an aqueous solution. 25.0 mL of solution is used when the HCl solution was titrated against it. The numerical value of is . (Nearest integer) Given : Molar mass of HCl and are 36.5 and respectively.
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Show full solutionCorrect answer: 1825
Correct answer
1825
Step-by-step explanation
The reaction between HCl and Ba(OH)₂ is:
2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
Moles of Ba(OH)₂ used: n = 0.1 M × 0.025 L = 0.0025 mol
From stoichiometry: moles of HCl = 2 × 0.0025 = 0.005 mol
Mass of HCl = 0.005 mol × 36.5 g/mol = 0.1825 g = 182.5 mg
Expressing as x = ___ × 10⁻¹:
182.5 = 1825 × 10⁻¹
The numerical value is 1825 (or 1.825 × 10² mg when expressed with standard scientific notation)
2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
Moles of Ba(OH)₂ used: n = 0.1 M × 0.025 L = 0.0025 mol
From stoichiometry: moles of HCl = 2 × 0.0025 = 0.005 mol
Mass of HCl = 0.005 mol × 36.5 g/mol = 0.1825 g = 182.5 mg
Expressing as x = ___ × 10⁻¹:
182.5 = 1825 × 10⁻¹
The numerical value is 1825 (or 1.825 × 10² mg when expressed with standard scientific notation)
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