x\, mg of pure HCl was used to make an aqueous solution. 25.0\, mL of 0.1\, M Ba(OH) 2 solution is used when the HCl solution was titrated against it. The numerical value of x is \ \ \ \ × 10 -1 (nearest integer). (Given: molar mass of HCl =36.5 and Ba(OH) 2 =171.0\, g\,mol -1 )

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2026 — 23 January, Morning Shift — Question 68

$x\,\mathrm{mg}$ of pure $\mathrm{HCl}$ was used to make an aqueous solution. $25.0\,\mathrm{mL}$ of $0.1\,\mathrm{M}$ $\mathrm{Ba(OH)_2}$ solution is used when the $\mathrm{HCl}$ solution was titrated against it. The numerical value of $x$ is $\_\_\_\_ \times 10^{-1}$ (nearest integer).

(Given: molar mass of $\mathrm{HCl}=36.5$ and $\mathrm{Ba(OH)_2}=171.0\,\mathrm{g\,mol^{-1}}$ )

Answer: 1825

Numerical answer — enter this value.

Step-by-step solution

Balanced reaction: $\mathrm{Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O}$

Moles of $\mathrm{Ba(OH)_2}$ $= 0.1 \times \frac{25.0}{1000} = 0.0025\,\mathrm{mol}$

From stoichiometry, $1\,\mathrm{Ba(OH)_2} \rightarrow 2\,\mathrm{HCl}$

Moles of $\mathrm{HCl}$ required $= 2 \times 0.0025 = 0.005\,\mathrm{mol}$

Mass of $\mathrm{HCl}$ $= 0.005 \times 36.5 = 0.1825\,\mathrm{g}$ $= 182.5\,\mathrm{mg}$

Thus, $x = 182.5 = 1825 \times 10^{-1}$

Nearest integer = $1825$

Answer key and solution verified before publishing.

Practise Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Chemical Equations, Stoichiometry and Limiting Reagent