Given:
- Freezing point of a 4% aqueous solution of X is equal to the freezing point of the 12% aqueous solution of Y.
- Molecular weight of X is A.

To find:
- Molecular weight of Y.

Solution:
The freezing point of a solution is directly related to the concentration of solute particles in the solution. According to the colligative property of freezing point depression, the freezing point of a solution is lowered when a non-volatile solute is added to it.

Step 1: Understand the problem
We are given two solutions, one containing 4% of X and another containing 12% of Y. We need to find the molecular weight of Y.

Step 2: Express the given information mathematically
The freezing point depression is directly proportional to the molality of the solution. The formula for freezing point depression is:

ΔT = Kf * m

Where:
ΔT = Change in freezing point
Kf = Cryoscopic constant of the solvent
m = Molality of the solution

Since the freezing point of the two solutions is the same, we can equate the freezing point depressions:

Kf * m1 = Kf * m2

Step 3: Substitute the given values
Let's assume the molecular weight of Y is B. Therefore, the molality of the 12% solution of Y can be calculated as follows:

m2 = (12% / B)

The molality of the 4% solution of X can be calculated as follows:

m1 = (4% / A)

Substituting these values into the equation:

Kf * (4% / A) = Kf * (12% / B)

Step 4: Simplify the equation
Since the cryoscopic constant and the solvent are the same for both solutions, we can cancel them out:

4% / A = 12% / B

Step 5: Solve for B
To find the value of B, we need to isolate it on one side of the equation. We can do this by cross-multiplying:

4% * B = 12% * A

Simplifying further:

0.04 * B = 0.12 * A

Dividing both sides by 0.04:

B = (0.12 * A) / 0.04

Simplifying:

B = 3A

Therefore, the molecular weight of Y is 3 times the molecular weight of X, which can be represented as A. Thus, the correct answer is option 'A' (3A).