Question

Calculate the elevation of boiling point of a solution when 3 g of CaCl2 (Molar mass = 111 g mol\textsuperscript{-1}) was dissolved in 260 g of water, assuming that CaCl2 undergoes complete dissociation. (Kb for water = 0.52 K kg mol\textsuperscript{-1})

Hint:

For colligative properties like boiling point elevation, remember that the van 't Hoff factor (\(i\)) accounts for the number of particles formed in solution.

Answer 1

The elevation of boiling point (\(\Delta T_b\)) is given by the formula: \[ \Delta T_b = i \cdot K_b \cdot m \] where \(i\) is the van 't Hoff factor, \(K_b\) is the ebullioscopic constant, and \(m\) is the molality of the solution.
For CaCl2, \(i = 3\) (since it dissociates into 3 ions: Ca2+ and 2Cl\textsuperscript{-}).
Molality \(m\) = \(\dfrac{moles\ of\ solute}{kg\ of\ solvent} = \dfrac{3}{111 \times 0.260} = 0.103\ mol\ kg^{-1}\)
\[ \Delta T_b = 3 \times 0.52\ \text{K kg mol}^{-1} \times 0.103\ \text{mol kg}^{-1} = 0.161\ K \] Thus, the elevation in boiling point is 0.161 K.