1. Calculate the freezing point of a 0.2 m CaCl2 aqueous solution, assuming it dissociates completely to form ions in solution
i for CaCl2 has it dissociates into 3 particles. [1 Ca2+ and 2 Cl-]
lets now calculate ΔTf
ΔTf = i*Kf*m
= 3.0*1.86*0.2
= 1.116 oC
This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 1.116
= -1.116 oC
Answer: -1.12 oC
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