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

Chemistry · p-Block Elements (Group 15-18)

JEE Main 2026 — 24 January, Evening Shift — Question 55

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

Option A:
$0.75 \mathrm{M}, 0.75 \mathrm{M}, 1 \mathrm{M}$
Option B:
$0.5 \mathrm{M}, 0.5 \mathrm{M}, 0.5 \mathrm{M}$
Option C:
$0.5 \mathrm{M}, 0.5 \mathrm{M}, 1 \mathrm{M}$
Correct
Option D:
$1 \mathrm{M}, 1 \mathrm{M}, 1 \mathrm{M}$

Answer: C

Step-by-step solution

$\mathrm{Cl}_{2}+2 \mathrm{KOH} \rightarrow \mathrm{KCl}+\mathrm{KClO}+\mathrm{H}_{2} \mathrm{O}$ $\mathrm{t}=01$ mole 4 mole $\mathrm{t}_{\mathrm{f}} 02$ mole 1 mole 1 mole $\left[\mathrm{OH}^{-}\right]=1 \mathrm{M}$ $\left[\mathrm{Cl}^{-}\right]=\frac{1}{2} \mathrm{M}$ $\left[\mathrm{ClO}^{-}\right]=\frac{1}{2} \mathrm{M}$

Answer key and solution verified before publishing.

Practise p-Block Elements (Group 15-18)

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: 24 January, Evening Shift
Subject: Chemistry
Chapter: p-Block Elements (Group 15-18)
Topic: Compounds of Halogens