JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Practical and Analytical Chemistry

x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M

Ba (OH) X_{2}

solution is used when the HCl solution was titrated against it. The numerical value of x is ____

\times 1 0^{- 1}

. (Nearest integer)
Given : Molar mass of HCl and

Ba (OH) X_{2}

are 36.5 and 171.0 g mol

^{- 1}

respectively.

Answer: 1825

Solution

Neutralisation follows

Ba(OH)_{2} + 2 HCl \to BaCl_{2} + 2 H_{2} O

. The base supplies

n_{Ba(OH)_{2}} = 0.1 \times \frac{25.0}{1000} = 2.5 \times 1 0^{- 3} mol

, and each mole neutralises two moles of acid, so

n_{HCl} = 5.0 \times 1 0^{- 3} mol

. The mass of HCl is

5.0 \times 1 0^{- 3} \times 36.5 = 0.1825 g = 182.5 mg

. Since the mass is reported as

x \times 1 0^{- 1} mg

, the value of

x

1825

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Integer answer

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