JEE Chemistry 2026 Question with Solution

Given: $1$ mole of $\text{Cl}_2$ is passed into $2 \, \text{L}$ of cold $2 \, \text{M}$ KOH solution.

Find: The final concentrations of $\text{Cl}^-$, $\text{ClO}^-$ and $\text{OH}^-$.

For cold dilute alkali, chlorine forms hypochlorite.

Initial moles of KOH are:

So, initial moles of $\text{OH}^-$ are $4$.

From the reaction, $1$ mole of $\text{Cl}_2$ consumes $2$ moles of $\text{OH}^-$. Therefore, moles of $\text{OH}^-$ left are:

Also, from the stoichiometry:

Volume remains constant at $2 \, \text{L}$.

Hence,

Therefore, the concentrations are $0.5 \, \text{M}$, $0.5 \, \text{M}$, and $1 \, \text{M}$, so the correct option is C.

Given: Cold dilute KOH reacts with chlorine, and the solution volume is constant.

Find: Which option matches the final ionic concentrations.

The key idea is disproportionation of chlorine in cold alkali:

This means one chlorine atom is reduced to $\text{Cl}^-$ and the other is oxidized to $\text{ClO}^-$, so $1$ mole of $\text{Cl}_2$ gives $1$ mole each of $\text{Cl}^-$ and $\text{ClO}^-$.

The KOH present initially is:

Thus, $4$ moles of $\text{OH}^-$ are available. Since the reaction consumes $2$ moles of $\text{OH}^-$ per mole of $\text{Cl}_2$, the leftover $\text{OH}^-$ is $2$ moles.

Now divide each final mole value by $2 \, \text{L}$ to get concentration.

Therefore, the correct option is C.