Calculate the cryscopic constant of a solvent, If its latent heat of fusion is `180.75 kg mol^(-1)` and freezing point `16.6^@C`.

Answer

To calculate the cryoscopic constant (Kf) of a solvent, we can use the formula:

\[ K_f = \frac{R \cdot T_f^2}{1000 \cdot L_f} \]

Where:
- \( R \) is the universal gas constant (8.314 J/(mol·K))
- \( T_f \) is the freezing point of the solvent in Kelvin
- \( L_f \) is the latent heat of fusion (given as 180.75 kg mol\(^{-1}\))

### Step-by-Step Solution:

1. **Convert the Freezing Point to Kelvin:**
\[ T_f = 16.6^\circ C + 273.15 = 289.75 \, K \]

2. **Square the Freezing Point:**
\[ T_f^2 = (289.75)^2 = 83970.0625 \, K^2 \]

3. **Substitute the Values into the Formula:**
\[ K_f = \frac{8.314 \cdot 83970.0625}{1000 \cdot 180.75} \]

4. **Calculate the Numerator:**
\[ 8.314 \cdot 83970.0625 = 6981550.4575 \]

5. **Calculate the Denominator:**
\[ 1000 \cdot 180.75 = 180750 \]

6. **Calculate Kf:**
\[ K_f = \frac{6981550.4575}{180750} \approx 38.66 \, K \cdot kg \cdot mol^{-1} \]

### Final Answer:
The cryoscopic constant \( K_f \) of the solvent is approximately \( 38.66 \, K \cdot kg \cdot mol^{-1} \).