Answer:
2,347.8 grams
Explanation:
The freezing point depression Kf of water = 1.86° C / molal.
To still freeze at -12° C, then the molality of the solution 12/ 1.86 = 6.45 moles
The molecular weight of sorbitol (C6H14O6)is:
6 C = 6 ×12 = 72
14 H = 14 × 1 = 14
6 O = 6 × 16 = 96
...giving a total of 182
So one mole of sorbitol has a mass of 182 grams.
Since there are 2 kg of water, 2 × 6.45 moles = 12.9 moles can be added to the water to get the 12° C freezing point depression.
Therefore
grams = moles × molar mass
12.9 moles × 182 grams / mole = 2,347.8 grams of sorbitol can be added and still freeze
