delta T = i*Kf*molality
i = 2.7
k = 1.86
molality = mols/kg solution. mols = 6.60/molar mass MgCl2 and kg solvent = 0.110
Finally, 0-delta T = new freezing point.
What is the freezing point in ∘C of a solution prepared by dissolving 6.60 g of MgCl2 in 110 g of water? The value of Kf for water is 1.86 (∘C⋅kg)/mol, and the van't Hoff factor for MgCl2 is i = 2.7.
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