van't Hoff factor of 0.1 molal `CaCl_2` solution will be (Assume `CaCl_2` dissociates completely in solution)

Options

A. 2
B. 4
C. 1
D. 3

Answer

To find the van't Hoff factor (i) for a 0.1 molal solution of \( \text{CaCl}_2 \), we need to follow these steps:

### Step 1: Understand the dissociation of \( \text{CaCl}_2 \)
When \( \text{CaCl}_2 \) dissolves in water, it dissociates completely into its constituent ions:
\[ \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^- \]
This means that one formula unit of \( \text{CaCl}_2 \) produces one calcium ion and two chloride ions.

### Step 2: Count the total number of particles formed
From the dissociation:
- 1 mole of \( \text{CaCl}_2 \) produces:
- 1 mole of \( \text{Ca}^{2+} \)
- 2 moles of \( \text{Cl}^- \)

Thus, the total number of moles of ions produced from 1 mole of \( \text{CaCl}_2 \) is:
\[ 1 + 2 = 3 \text{ moles of ions} \]

### Step 3: Determine the initial number of particles
Initially, before dissociation, we have:
- 1 mole of \( \text{CaCl}_2 \)

### Step 4: Calculate the van't Hoff factor (i)
The van't Hoff factor \( i \) is defined as the ratio of the number of particles in solution after dissociation to the number of particles before dissociation:
\[ i = \frac{n_{\text{final}}}{n_{\text{initial}}} \]
Substituting the values we found:
\[ i = \frac{3 \text{ moles of ions}}{1 \text{ mole of } \text{CaCl}_2} = 3 \]

### Conclusion
The van't Hoff factor \( i \) for a 0.1 molal \( \text{CaCl}_2 \) solution is 3.

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