The value of observed and calculated molecular weight of calcium nitrate are respectively 65.6 and 164. The degree of dissociation of calcium nitrate is
a) 25%
b) 50%
c) 75%
d) 60%

The question asks for the degree of dissociation (\(\alpha\)) of calcium nitrate, given that the observed and calculated molecular weights are 65.6 and 164, respectively.

### Step-by-step solution:

1. **Understanding dissociation and Van't Hoff factor:**
When a substance dissociates in solution, the effective number of particles in the solution changes. This is represented by the Van't Hoff factor (\(i\)), which is related to the degree of dissociation (\(\alpha\)).

Calcium nitrate \(\text{Ca(NO}_3\text{)}_2\) dissociates as:
\[
\text{Ca(NO}_3\text{)}_2 \rightarrow \text{Ca}^{2+} + 2\text{NO}_3^-
\]
Therefore, 1 molecule of calcium nitrate dissociates into 3 particles (1 calcium ion and 2 nitrate ions). Hence, the theoretical Van’t Hoff factor \(i_{\text{theoretical}}\) is 3.

2. **Van't Hoff factor formula:**
The observed Van’t Hoff factor (\(i_{\text{observed}}\)) can be calculated from the observed and calculated molecular weights using the relation:
\[
i_{\text{observed}} = \frac{\text{Calculated Molecular Weight}}{\text{Observed Molecular Weight}}
\]
Substituting the given values:
\[
i_{\text{observed}} = \frac{164}{65.6} = 2.5
\]

3. **Relating Van’t Hoff factor to degree of dissociation:**
The degree of dissociation (\(\alpha\)) is related to the Van’t Hoff factor by the formula:
\[
i_{\text{observed}} = 1 + \alpha(n - 1)
\]
where:
- \(i_{\text{observed}}\) is the observed Van’t Hoff factor,
- \(\alpha\) is the degree of dissociation,
- \(n\) is the number of particles formed after dissociation (in this case, \(n = 3\)).

Substituting \(i_{\text{observed}} = 2.5\) and \(n = 3\):
\[
2.5 = 1 + \alpha(3 - 1)
\]
\[
2.5 = 1 + 2\alpha
\]
\[
2\alpha = 2.5 - 1 = 1.5
\]
\[
\alpha = \frac{1.5}{2} = 0.75
\]

4. **Conclusion:**
The degree of dissociation (\(\alpha\)) is 0.75, or 75%.

Thus, the correct answer is **(c) 75%**.