Question

If latent heat of fusion of ice is 80 cals per g at $0^\circ C$, calculate molal depression constant for water

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

Answer 1

Answer: (C)

1.863

Answer 2

The molal depression constant ($K_f$) can be calculated using the following formula:

$K_f = \frac{R T_f^2}{L_f \times 1000}$

Where:

• $R$ is the Universal Gas Constant.
• $T_f$ is the freezing point of the solvent in Kelvin.
• $L_f$ is the latent heat of fusion of the solvent per gram.

Given values:

1. Latent heat of fusion of ice ($L_f$) = 80 cals/g
2. Freezing point of water ($T_f$) = $0^\circ C$. To convert to Kelvin, add 273: $T_f = 0 + 273 = 273 \text{ K}$.
3. Since the latent heat is given in calories, we use the value of the Universal Gas Constant ($R$) in calories: $R = 2 \text{ cal/mol K}$.

Now, substitute these values into the formula:

$K_f = \frac{2 \text{ cal/mol K} \times (273 \text{ K})^2}{80 \text{ cal/g} \times 1000 \text{ g/kg}}$ $K_f = \frac{2 \times (273)^2}{80 \times 1000}$ $K_f = \frac{2 \times 74529}{80000}$ $K_f = \frac{149058}{80000}$ $K_f = 1.863225 \text{ K kg/mol}$

Rounding to three decimal places, $K_f = 1.863 \text{ K kg/mol}$.