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)

We need to find the concentrations of Cl⁻, ClO⁻, and OH⁻ after passing Cl₂ into cold KOH.

$$Cl_2 + 2KOH \rightarrow KCl + KClO + H_2O$$

1 mol Cl₂ added to 2 L of 2M KOH (total KOH = 4 mol).

Stoichiometry: 1 mol Cl₂ reacts with 2 mol KOH.

Products formed:

KCl: 1 mol → Cl⁻ = 1 mol

KClO: 1 mol → ClO⁻ = 1 mol

KOH consumed: 2 mol

KOH remaining: 4 - 2 = 2 mol → OH⁻ = 2 mol

Concentrations (in 2 L):

[Cl⁻] = 1/2 = 0.5 M

[ClO⁻] = 1/2 = 0.5 M

[OH⁻] = 2/2 = 1 M

Therefore, the answer is Option 1: 0.5M, 0.5M, 1M.