If latent heat of fusion of ice is `80` cals per g at `0^(@)`, calculate molal depression constant for water.

Options

A. 18.63
B. 186.3
C. 1.863
D. 0.1863

Answer

To calculate the molal depression constant (Kf) for water using the given latent heat of fusion of ice, we can follow these steps:

### Step 1: Understand the given data
The latent heat of fusion of ice is given as 80 calories per gram at 0°C (or 273 K).

### Step 2: Convert the latent heat of fusion to molar terms
To find the molar heat of fusion, we need to convert the latent heat from calories per gram to calories per mole. The molar mass of water (H2O) is approximately 18 g/mol.

\[ \text{Molar heat of fusion} = 80 \, \text{cal/g} \times 18 \, \text{g/mol} = 1440 \, \text{cal/mol} \]

### Step 3: Use the formula for molal depression constant
The formula for the molal depression constant (Kf) is given by:

\[ K_f = \frac{R T^2}{1000 \cdot \Delta H_{fusion}} \]

Where:
- \( R \) is the gas constant (2 cal/K·mol for our calculations),
- \( T \) is the temperature in Kelvin (273 K),
- \( \Delta H_{fusion} \) is the molar heat of fusion (1440 cal/mol).

### Step 4: Substitute the values into the formula
Now we can substitute the values into the formula:

\[ K_f = \frac{2 \, \text{cal/K·mol} \times (273 \, \text{K})^2}{1000 \times 1440 \, \text{cal/mol}} \]

### Step 5: Calculate \( K_f \)
First, calculate \( (273)^2 \):

\[ (273)^2 = 74529 \]

Now substitute this back into the equation:

\[ K_f = \frac{2 \times 74529}{1000 \times 1440} \]

Calculating the numerator:

\[ 2 \times 74529 = 149058 \]

Now calculate the denominator:

\[ 1000 \times 1440 = 1440000 \]

Now divide the numerator by the denominator:

\[ K_f = \frac{149058}{1440000} \approx 0.1035 \, \text{cal/K·kg} \]

### Final Result
Thus, the molal depression constant \( K_f \) for water is approximately **0.1035 cal/K·kg**.

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