Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y .If molecular weight of X is A then molecular weight of Y is :

Options

A. 3A
B. A
C. 4A
D. 2A

Explanation

To solve the problem, we need to find the molecular weight of substance Y in terms of the molecular weight of substance X (denoted as A). The key to this problem lies in the concept of molality and the fact that the freezing points of the two solutions are equal.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two solutions: a 4% aqueous solution of X and a 12% aqueous solution of Y. The freezing points of both solutions are equal, which implies that their molalities are also equal.

2. **Calculating Masses**:
- For the 4% solution of X:
- Total mass of the solution = 100 g
- Mass of X = 4 g (4% of 100 g)
- Mass of water (solvent) = 100 g - 4 g = 96 g

- For the 12% solution of Y:
- Total mass of the solution = 100 g
- Mass of Y = 12 g (12% of 100 g)
- Mass of water (solvent) = 100 g - 12 g = 88 g

3. **Calculating Molality**:
- Molality (m) is defined as the number of moles of solute per kilogram of solvent.
- For X:
\[ \text{Number of moles of X} = \frac{4 \text{ g}}{M_X} \quad \text{(where } M_X \text{ is the molecular weight of X)} \]
\[ \text{Mass of solvent (water)} = 96 \text{ g} = 0.096 \text{ kg} \]
\[ m_X = \frac{\frac{4}{M_X}}{0.096} = \frac{4 \times 1000}{M_X \times 96} \]

- For Y:
\[ \text{Number of moles of Y} = \frac{12 \text{ g}}{M_Y} \quad \text{(where } M_Y \text{ is the molecular weight of Y)} \]
\[ \text{Mass of solvent (water)} = 88 \text{ g} = 0.088 \text{ kg} \]
\[ m_Y = \frac{\frac{12}{M_Y}}{0.088} = \frac{12 \times 1000}{M_Y \times 88} \]

4. **Setting Molalities Equal**:
Since the freezing points are equal, we set the molalities equal:
\[ \frac{4 \times 1000}{M_X \times 96} = \frac{12 \times 1000}{M_Y \times 88} \]

5. **Simplifying the Equation**:
Canceling out the 1000 from both sides:
\[ \frac{4}{M_X \times 96} = \frac{12}{M_Y \times 88} \]
Cross-multiplying gives:
\[ 4 \times 88 = 12 \times M_X \times 96 / M_Y \]
Rearranging for \(M_Y\):
\[ M_Y = \frac{12 \times 88}{4 \times 96} \times M_X \]

6. **Calculating Molecular Weight of Y**:
Simplifying the fraction:
\[ M_Y = \frac{12 \times 88}{4 \times 96} \times A \]
\[ = \frac{12 \times 88}{384} \times A = \frac{1056}{384} \times A = \frac{11}{4} \times A = 3A \]

### Final Answer:
The molecular weight of Y is \(3A\).