Question

Assume that the solubility of CaCl2 at this temperature is 70.1 g CaCl2/100.0 g of H2O and that the van't Hoff factor for a saturated solution of CaCl2 is i = 2.5. The Kf for water is 1.86°C/m What is the minimum amount of CaCl2 that would be required to melt ice on sidewalks at the above temperature- -33.0 degrees C? Use 100.0 g of water as the amount of solvent. Give your answer in grams.

Answer 1

Tf = i*Kf*m

depression constant (Kf),i=van't Hoff factor,Tf=final temp

molality, m = (70.1/111)/0.1 = 6.3153

Tf = 2.5 * 1.86*6.3153 = 29.366 Celcius

Hence, maximum it can lower freezing point up to -30 degree Celsius.

Therefore, NO, ice could not melt at -33 degrees Celsius

The theoretical amount of CaCl2 (which is not actually possible because of the solubility of CaCl2) is:
(33 °C change) / (1.86 °C/m) = 17.774 m ions
(17.774 m ions) / 2.5 = 7.096 m CaCl2
(7.096 mol CaCl2/kg) x (0.100 kg) x (110.984 g CaCl2/mol) = 78.76 g CaCl2
But this amount of CaCl2 will not dissolve in 100 g of water at that temperature.