What is the freezing point of a solution that is made by dissolving 20.0g of MgCl2 in 100.0g of...

Question

What is the freezing point of a solution that is made by dissolving 20.0g of MgCl2 in 100.0g of water?

physicsproblem · Answer

delta Tf = i X Kf X m   where i = vant hoff factor   MgCl2 ---------> Mg2+ + 2Cl-   i = no.of particles after dissociation = 1 + 2 = 3   Kf = 1.86 degreeC/m   m = molality = no.of moles of MgCl2/amount of water in kg   molecular mass of MgCl2 = 95 g/mole   so no.of moles of MgCl2 = 20/95 = 0.210  amount of water in kg = 100/1000 = 0.1 kg   so molality = 0.210/0.1 = 2.1 moles/kg   so delta Tf = 3 X 1.86 X 2.1 = 11.718  so freezing point of solution = 0 - 11.718 = -11.718 degree C

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What is the freezing point of a solution that is made by dissolving 20.0g of MgCl2 in 100.0g of...

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