Question

What is the freezing point (in C) of a 1.56 m aqueous solution of CaCl2? (Amount to three decimal points)

Answer 1

The freezing point of a 1.56 m aqueous solution of CaCl2, using freezing point depression, is calculated to be -29.6988°C.

Step-by-step explanation:

The freezing point of a solution can be determined using the concept of freezing point depression, which is a colligative property. For a 1.56 m aqueous solution of CaCl2, which is an ionic compound that dissociates in water, the freezing point depression can be calculated using the formula ΔTf = i * Kf * m.

ΔTf = change in freezing point (freezing point depression)
i = van't Hoff factor (number of particles the compound dissociates into in solution)
Kf = freezing point depression constant for the solvent (water has a Kf of 1.86°C kg/mol)
m = molality of the solution

Since CaCl2 dissociates into 3 ions (1 Ca2+ and 2 Cl–), the van't Hoff factor (i) is 3. Applying this to the formula:

ΔTf = 3 * 1.86°C kg/mol * 1.56 m = 8.6988°C

Therefore, the freezing point of the 1.56 m CaCl2 solution is 0°C - 8.6988°C = ‑29.6988°C.

Answer 2

Supposing complete ionization:
CaCl2 → Ca{2+} + 2 Cl{-} [three ions total]

(1.56 m CaCl2) x (3 mol ions / 1 mol CaCl2) = 4.68 m ions

(1.86 °C/m) x (4.68 m) = 8.70 °C change

0°C - 8.70°C = - 8.70°C