Calculate the molal depression constant of water. Latent heat of fusion of ice at `0^(@)` at `80` calories per gram

Answer

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

### Step 1: Understand the formula
The molal depression constant (Kf) can be calculated using the formula:
\[ K_f = \frac{R \cdot T^2}{1000 \cdot L_f} \]
where:
- \( R \) = gas constant (in calories per gram per Kelvin)
- \( T \) = absolute temperature (in Kelvin)
- \( L_f \) = latent heat of fusion (in calories per gram)

### Step 2: Identify the values
From the question, we have:
- Latent heat of fusion of ice, \( L_f = 80 \) calories/gram
- Temperature at freezing point, \( T = 0^\circ C = 273 \) K
- Gas constant, \( R = 2 \) calories/(gram·K)

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the formula:
\[ K_f = \frac{2 \cdot (273)^2}{1000 \cdot 80} \]

### Step 4: Calculate \( T^2 \)
First, calculate \( (273)^2 \):
\[ (273)^2 = 74529 \]

### Step 5: Substitute \( T^2 \) back into the equation
Now substitute \( T^2 \) back into the equation:
\[ K_f = \frac{2 \cdot 74529}{1000 \cdot 80} \]

### Step 6: Calculate the numerator
Calculate the numerator:
\[ 2 \cdot 74529 = 149058 \]

### Step 7: Calculate the denominator
Calculate the denominator:
\[ 1000 \cdot 80 = 80000 \]

### Step 8: Divide the numerator by the denominator
Now divide the numerator by the denominator:
\[ K_f = \frac{149058}{80000} \]

### Step 9: Perform the final calculation
Calculating the division gives:
\[ K_f \approx 1.863225 \]

### Step 10: Round off the answer
Rounding off to two decimal places, we get:
\[ K_f \approx 1.86 \, \text{°C} \]

### Final Answer
Thus, the molal depression constant of water is approximately:
\[ K_f \approx 1.86 \, \text{°C} \]

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