Question

Calculate the mass of CaCl$_2$ (molar mass = 111 g mol$^{-1}$) to be dissolved in 500 g of water to lower its freezing point by 2K, assuming that CaCl$_2$ undergoes complete dissociation.

Hint:

Freezing point depression depends on the molality and the Van’t Hoff factor. For ionic compounds like CaCl$_2$, dissociation increases the number of particles and the effect on the freezing point.

Answer 1

The freezing point depression formula is applied as follows:
\[\Delta T_f = i \times K_f \times m\]Given values are:
- \( \Delta T_f = 2 \, \text{K} \),
- \( K_f = 1.86 \, \text{K kg mol}^{-1} \),
- \( i = 3 \) (for CaCl$_2$, which dissociates into Ca$^{2+}$ and 2 Cl$^{-}$),
- \( m \) represents the molality of the solution.
Step 1: Solve for molality \( m \):
\[m = \frac{\Delta T_f}{i \times K_f} = \frac{2}{3 \times 1.86} = 0.358 \, \text{mol/kg}\]Step 2: Calculate the moles of CaCl$_2$ using the molality and the mass of the solvent (0.5 kg):
\[\text{moles of CaCl}_2 = m \times \text{mass of solvent} = 0.358 \times 0.5 = 0.179 \, \text{mol}\]Step 3: Calculate the mass of CaCl$_2$ using the moles and the molar mass (111 g/mol):
\[\text{mass of CaCl}_2 = \text{moles} \times \text{molar mass} = 0.179 \times 111 = 19.89 \, \text{g}\]Therefore, 19.89 g of CaCl$_2$ is required.