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In the winter of 1994, record low temperatures were registered throughout the United States. For example, in Champaign, Illinois, a record low of -29°F was registered. At this temperature can salting icy roads with CaCl2 be effective in melting the ice? (The solubility of CaCl2 in cold water is 74.5 g per 100.0 g of water.)
(a) Assume i = 3.00 for CaCl2.
(b) Assume I = 2.60

Why does it work for 3.00 but not 2.60? I'm confused as to how I process this question and explain it.

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MissPhiladelphia
Hi, thank you for posting your question here at Brainly.

Freezing point depression is one of the colligative properties. Its equation is

[tex] T_{F,solvent} - T_{F,solution} =molality*i* K_{F,solvent} [/tex]

where the solvent is water. Thus, TF,solvent = 0 degrees Celsius and Kf is 1.86 kg/mol-C. i is the vant Hoff factor and TF,solution is -29 degrees = -33.9 degrees Celsius. Substituting,

For i = 3,
[tex]0 - -33.9 =molality*3* 1.86[/tex]
molality = 6.08 mol  CaCl2/kg water

For i=2.6
[tex]0 - -33.9 =molality*2.6* 1.86[/tex]
molality = 7 mol CaCl2/kg water

We compare the values of the molalities to the solubility. Molecular weight of CaCl2 = 111 g/mol.

Solubility = (74.5 g/100g)(1 mol /111g)(1000 g/1 kg)
Solubility = 6.71 mol CaCl2/kg water

Since at 7 mol CaCl2/kg water (when i=2.6) is greater than the solubility, then it will not be possible. 

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