One use of CaCl2 is for salting roads in the winter. How much would the freezing point of water decrease if a 3.23 molal (m) solution of CaCl2 were applied? (Kf = 1.86°C/(mol/kg) for water and i = 3 for CaCl2.)

Freezing point of water will decrease to -18.02°C

Explanation:

We start from the formula for freezing point depression, one of the colligative properties.

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

m = molality

Kf = Cyoscopic constant (specific for each solvent)

i = numbers of ions dissolved (Van't Hoff factor)

CaCl₂ →  Ca²⁺  + 2Cl⁻    i = 3

We replace each data:

0°C - T° F) = 1.86°C/m . 3.23 m . 3

T°F = - 18.02°C

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