Equimolal solutions `KCl` and compound `X` in water show depression in freezing point in the ratio of `4:1`, Assuming `KCl` to be completely ionized, the compound `X` in solution must

Options

A. Dissociate to the extent of `50%`
B. Hydrolyze to the extent of `80%`
C. Dimerize to the extent of `50%`
D. Trimerize to the extent of `75%`

Explanation

To solve the problem, we need to analyze the depression in freezing point of equimolal solutions of KCl and compound X. We know that KCl completely ionizes in solution. Here’s the step-by-step solution:

### Step 1: Understand the relationship between depression in freezing point and van’t Hoff factor
The depression in freezing point (ΔTf) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) = depression in freezing point
- \( i \) = van’t Hoff factor (number of particles the solute breaks into)
- \( K_f \) = cryoscopic constant
- \( m \) = molality of the solution

### Step 2: Set up the ratio of depression in freezing point
We are given that the ratio of depression in freezing point for KCl to compound X is 4:1. Therefore, we can write:
\[ \frac{\Delta T_f(KCl)}{\Delta T_f(X)} = \frac{4}{1} \]

### Step 3: Substitute the formula for ΔTf
Using the formula for ΔTf, we can express this ratio as:
\[ \frac{i_{KCl} \cdot K_f \cdot m}{i_X \cdot K_f \cdot m} = \frac{4}{1} \]
Since \( K_f \) and \( m \) are the same for both solutions, they cancel out:
\[ \frac{i_{KCl}}{i_X} = 4 \]

### Step 4: Determine the van’t Hoff factor for KCl
KCl dissociates into two ions (K\(^+\) and Cl\(^-\)), so:
\[ i_{KCl} = 2 \]

### Step 5: Solve for the van’t Hoff factor for compound X
Substituting \( i_{KCl} \) into the equation gives:
\[ \frac{2}{i_X} = 4 \]
Cross-multiplying gives:
\[ 2 = 4 \cdot i_X \implies i_X = \frac{2}{4} = 0.5 \]

### Step 6: Interpret the van’t Hoff factor for compound X
A van’t Hoff factor of 0.5 indicates that the compound X is associating in solution. If we denote the association as:
\[ 2X \rightleftharpoons X_2 \]
This means that two molecules of X combine to form one molecule of \( X_2 \).

### Step 7: Determine the extent of association
Using the formula for association:
\[ i = 1 - \frac{1}{n} \cdot \alpha \]
where \( n \) is the number of particles formed from association (in this case, \( n = 2 \) because 2X forms 1 \( X_2 \)), we can substitute:
\[ 0.5 = 1 - \frac{1}{2} \cdot \alpha \]
Rearranging gives:
\[ \frac{1}{2} \cdot \alpha = 1 - 0.5 \implies \frac{1}{2} \cdot \alpha = 0.5 \implies \alpha = 1 \]

### Step 8: Calculate the degree of association
Since \( \alpha \) represents the fraction of molecules that remain unassociated, we find:
\[ \alpha = 1 - \frac{1}{2} \cdot \alpha \implies \alpha = 0.75 \]
This means that 75% of the molecules are associated.

### Conclusion
The compound X must be a substance that associates in solution to the extent of 75%, indicating that it forms dimers in solution.