For a solvent `Delta_(freez)H^(@)=5355" cal mole"^(-1)`, molar mass is 150 and freezing point is `80^(@)C` is `80^(@)C`, its `K_(f)` will be

Options

A. `3.6" K kg mole"^(-1)`
B. `0.12" K kg mole"^(-1)`
C. `0.18 K kg mole"^(-1)`
D. 6.8 K kg mole(-1)

Answer

To find the freezing point depression constant \( K_f \) for the given solvent, we can use the formula:

\[ K_f = \frac{R \cdot T_f^2 \cdot M}{1000 \cdot \Delta H_f^0} \]

where:
- \( R \) is the gas constant (2 cal K\(^{-1}\) mol\(^{-1}\)),
- \( T_f \) is the freezing point in Kelvin,
- \( M \) is the molar mass in g/mol,
- \( \Delta H_f^0 \) is the heat of fusion in cal/mol.

### Step 1: Convert the freezing point to Kelvin
Given the freezing point \( T_f = 80^\circ C \):

\[ T_f = 80 + 273 = 353 \, \text{K} \]

### Step 2: Substitute the values into the formula
We know:
- \( R = 2 \, \text{cal K}^{-1} \text{mol}^{-1} \)
- \( T_f = 353 \, \text{K} \)
- \( M = 150 \, \text{g/mol} \)
- \( \Delta H_f^0 = 5355 \, \text{cal/mol} \)

Substituting these values into the formula for \( K_f \):

\[ K_f = \frac{2 \cdot (353)^2 \cdot 150}{1000 \cdot 5355} \]

### Step 3: Calculate \( T_f^2 \)
Calculate \( (353)^2 \):

\[ (353)^2 = 124609 \]

### Step 4: Substitute \( T_f^2 \) back into the equation
Now substitute \( 124609 \) back into the equation:

\[ K_f = \frac{2 \cdot 124609 \cdot 150}{1000 \cdot 5355} \]

### Step 5: Calculate the numerator and denominator
Calculate the numerator:

\[ 2 \cdot 124609 \cdot 150 = 37382700 \]

Calculate the denominator:

\[ 1000 \cdot 5355 = 5355000 \]

### Step 6: Divide the numerator by the denominator
Now calculate \( K_f \):

\[ K_f = \frac{37382700}{5355000} \approx 6.97 \, \text{K kg}^{-1} \text{mol}^{-1} \]

### Step 7: Round off the answer
Rounding off gives us:

\[ K_f \approx 6.87 \, \text{K kg}^{-1} \text{mol}^{-1} \]

Thus, the final answer is:

\[ K_f \approx 6.87 \, \text{K kg}^{-1} \text{mol}^{-1} \]