In reaction of 

With $KOH/I_2$  to form Iodoform. How many mole of KOH are consumed per mole of Ketone
 

We are talking about the Iodoform reaction:

Reaction: Methyl ketones (like CH₃CO-R) react with I₂ and KOH to give CHI₃ (iodoform) and a carboxylate.

Step 1: How it works

The methyl group (–CH₃) next to the carbonyl (C=O) is halogenated three times by iodine in alkaline KOH.

Then, the –CI₃ group is split off, leaving a carboxylate (RCOO⁻).

Step 2: KOH consumption

For each methyl ketone, 3 I₂ molecules are needed → 3 moles of I₂.

Each halogenation step consumes 1 mole of OH⁻ → 3 moles of KOH for the halogenation.

Plus 1 extra mole of KOH to neutralize the carboxylic acid formed → total 4 moles of KOH per mole of ketone.

Step 3: Short summary

CH₃CO-R + 3 I₂ + 4 KOH → CHI₃ + RCOOK + 3 KI + 3 H₂O
CH₃CO-R + 3 I₂ + 4 KOH → CHI₃ + RCOOK + 3 KI + 3 H₂O

Answer: 4 moles of KOH per mole of ketone.