The reaction of 50% aq KOH on an equimolar mixture of 4-methylbenzaldehyde and formaldehyde followed by acidification gives.

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Answer

To solve the problem, we need to analyze the reaction between 4-methylbenzaldehyde and formaldehyde when treated with 50% aqueous KOH followed by acidification.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have 4-methylbenzaldehyde (C6H5CHO) and formaldehyde (HCHO) in an equimolar mixture.
- Structure of 4-methylbenzaldehyde: It has a benzene ring with a formyl group (CHO) and a methyl group (CH3) at the para position.

2. **Reaction with 50% Aqueous KOH**:
- The reaction with KOH suggests that we are performing a condensation reaction, specifically a benzoin condensation or similar reaction.
- Under basic conditions, the aldehyde groups can undergo nucleophilic addition.

3. **Formation of the Product**:
- The 4-methylbenzaldehyde will react with formaldehyde. The aldehyde group of formaldehyde can act as a nucleophile and attack the carbonyl carbon of 4-methylbenzaldehyde.
- The reaction leads to the formation of a new compound where the carbonyl group (C=O) of 4-methylbenzaldehyde is converted to a hydroxyl group (C-OH), resulting in a secondary alcohol.

4. **Acidification**:
- After the reaction with KOH, the mixture is acidified. Acidification will protonate the hydroxyl group, stabilizing the product.
- The final product after acidification will be a hydroxy compound derived from the original aldehydes.

5. **Final Product**:
- The final product of this reaction is 4-methylbenzyl alcohol (C6H5CH(OH)CH3) and formic acid (HCOOH) from the formaldehyde.
- Thus, the overall reaction yields 4-methylbenzyl alcohol and formic acid.

### Conclusion:
The reaction of 50% aqueous KOH on an equimolar mixture of 4-methylbenzaldehyde and formaldehyde followed by acidification gives 4-methylbenzyl alcohol and formic acid.