CaCl2 is often used to melt ice on sidewalks. Could CaCl2 melt ice at -31 degrees Celsius? Assume that the solubility of CaCl2 at this temp is 70.1 g CaCl2/100 g of H2O and that the van’t Hoff factor for a saturated solution of CaCl2 is i=2.5. What is the minimum amount of CaCl2 that would be required to melt ice on sidewalks at the above temperature? Use 100 g of water as the amount of solvent. Answer must be in grams
(70.1 g CaCl2) / (110.984 g CaCl2/mol) / (100 g) x (1000 g) = 6.3162 m
(1.86 °C/m) x (6.3162 m) x 2.5 = 29.37 °C change
So CaCl2 cannot protect sidewalks against freezing below −29.37 °C.
The theoretical amount of CaCl2 (which is not actually possible because of the solubility of CaCl2) is:
(31 °C change) / (1.86 °C/m) = 16.667 m ions
(16.667 m ions) / 2.5 = 6.6667 m CaCl2
(6.6667 mol CaCl2/kg) x (0.100 kg) x (110.984 g CaCl2/mol) = 74 g CaCl2
But this amount of CaCl2 will not dissolve in 100 g of water at that temperature.