JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

One mole of $Cl_{2} (g)$ was passed into 2 L of cold 2M KOH solution. After the reaction, the concentrations of $Cl^{-}$ , $ClO^{-}$ and $OH^{-}$ are respectively (assume volume remains constant)

A
0.75 M, 0.75 M, 1 M
B
0.5 M, 0.5 M, 0.5 M
C
0.5 M, 0.5 M, 1 M
D
1 M, 1M, 1 M

Solution

Cold dilute KOH reaction:

$Cl_{2} + 2 KOH \to KCl + KClO + H_{2} O$

Initial: 1 mole $Cl_{2}$ , 4 moles KOH (= 2 L $\times$ 2 M).

1 mole $Cl_{2}$ consumes 2 moles KOH and produces 1 mole each of $KCl$ and $KClO$ . Remaining KOH = $4 - 2 = 2$ moles.

In 2 L: $[Cl^{-}] = \frac{1}{2} = 0.5$ M, $[ClO^{-}] = \frac{1}{2} = 0.5$ M, $[OH^{-}] = \frac{2}{2} = 1$ M.

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