Question

The molar depression constant of water (\(K_f\)) is 1.86 K kg mol\(^{-1}\). What is the approximate \( \Delta H_{fus} \) (in kJ mol\(^{-1}\)) of water if it freezes at 273 K?

• A. \(3\)
• B. \(6\)
• C. \(9\)
• D. \(12\)

Hint:

Remember to convert units appropriately when working with thermodynamic equations to ensure the correct calculations of enthalpy changes.

Answer 1

The Correct Option is B

Step 1: Use the relation between molar enthalpy of fusion and cryoscopic constant. The formula relating the enthalpy of fusion (\(\Delta H_f\)) to the cryoscopic constant (\(K_f\)) is: \[ \Delta H_f = \frac{R T^2 K_f}{1000} \] where: - \( R = 8.314 \) J mol\(^{-1}\)K\(^{-1}\) (universal gas constant), - \( T = 273 \) K (freezing point of water), - \( K_f = 1.86 \) K kg mol\(^{-1}\). 
Step 2: Substitute the values into the equation. \[ \Delta H_f = \frac{(8.314) \times (273)^2 \times (1.86)}{1000} \] \[ \Delta H_f = \frac{8.314 \times 74529 \times 1.86}{1000} \] \[ \Delta H_f = \frac{1155152.5}{1000} = 6.1 { kJ/mol} \] Approximating, we get 6 kJ/mol. Thus, the correct answer is 6 kJ/mol.