The values of observed and calculated molecular weights of calcium nitrate are respectively `65.6` and 164. The degree of dissociation of calcium nitrate will be:

Options

A. (a) `25%`
B. (b) `50%`
C. (c ) `75%`
D. (d) `60%`

Explanation

To find the degree of dissociation (α) of calcium nitrate (Ca(NO₃)₂), we can use the relationship between the observed molecular weight and the calculated molecular weight.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Observed molecular weight (M_observed) = 65.6 g/mol
- Calculated molecular weight (M_calculated) = 164 g/mol

2. **Understand the Concept of Degree of Dissociation:**
- Calcium nitrate dissociates in solution as follows:
\[ \text{Ca(NO}_3\text{)}_2 \rightarrow \text{Ca}^{2+} + 2\text{NO}_3^{-} \]
- For every mole of calcium nitrate that dissociates, it produces 1 mole of Ca²⁺ and 2 moles of NO₃⁻, leading to a total of 3 moles of ions.

3. **Calculate the Van 't Hoff Factor (i):**
- The Van 't Hoff factor (i) is defined as:
\[ i = \frac{M_{\text{calculated}}}{M_{\text{observed}}} \]
- Substituting the values:
\[ i = \frac{164}{65.6} \approx 2.5 \]

4. **Relate the Van 't Hoff Factor to Degree of Dissociation:**
- The relationship between the degree of dissociation (α) and the Van 't Hoff factor (i) for calcium nitrate is given by:
\[ i = 1 + 2\alpha \]
- Here, 1 accounts for the undissociated calcium nitrate, and 2α accounts for the two nitrate ions produced.

5. **Set Up the Equation:**
- From the previous step, we can set up the equation:
\[ 2.5 = 1 + 2\alpha \]

6. **Solve for α:**
- Rearranging the equation gives:
\[ 2\alpha = 2.5 - 1 \]
\[ 2\alpha = 1.5 \]
\[ \alpha = \frac{1.5}{2} = 0.75 \]

7. **Convert α to Percentage:**
- To express the degree of dissociation as a percentage:
\[ \text{Degree of dissociation} = \alpha \times 100 = 0.75 \times 100 = 75\% \]

### Final Answer:
The degree of dissociation of calcium nitrate is **75%**.