JEE Main 202623 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 23 January
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Ionic Equilibrium
JEE Main 2026 · 23 January · Morning Shift · Chemistry · Ionic Equilibrium
x mg of pure HCl was used to make an aqueous solution. 25.0 mL of...
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.
Correct answer
0
Step-by-step solution
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)