A solution is prepared by dissolving 104.1 grams of barium chloride, BaCl2 in 4000 grams of water...?

what is the freezing point for this solution?

-0.70oC
-186oC
-3.72oC
-5.58oC

Answer:
BaCl2 MW= 208
104.1g = 0.5 mole
total moles of ions = 3x0.5=1.5 (Ba++ and Cl- and Cl-)
fp depression = 1.86 x 1.5 x (1000/4000)
fp = -0.70 deg C

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