JEE Main 2026 — Jan 23, Shift 1 — Chemistry solution

JEE Main2026Jan 23, Shift 1Chemistry

$x$ 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 ______ $\times 1 0^{- 1}$ . (Nearest integer)

Given : Molar mass of HCl and Ba(OH) $_{2}$ are $36.5$ and $171.0$ g mol $^{- 1}$ respectively.

Your answer

Solution

Reaction: Ba(OH) $_{2}$ + 2HCl → BaCl $_{2}$ + 2H $_{2}$ O.

Moles of Ba(OH) $_{2}$ $= 25.0 mL \times 0.1 M = 2.5$ mmol.

Moles of HCl required $= 2 \times 2.5 = 5.0$ mmol.

Mass of HCl $= 5.0 \times 36.5 = 182.5$ mg.

Given $x \times 1 0^{- 1} = 182.5$ mg, so $x = 1825$ .

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