A. glycerol, C3H5(HO)3 is a nonvolitile molecular liquid that is very soluable i

Question

A. glycerol, C3H5(HO)3 is a nonvolitile molecular liquid that is very soluable in water. A solution is made by dissolving 75.0 g of glycerol in 350 g of water. What is the freezing point of the solution?
Remember: Delta Tf=i x Kf x m where i is the vant Hoff factor and Kf= 1.86 degrees celcius/m.

Explanation / Answer

he equation for freezing point depression is delta Tf = i x Kf x molality of solute where delta Tf is the amout the freezing point is lowered in degrees C, i is the van't Hoff factor, and Kf is the molal freezing point constant for the solvent (water) = 1.86 C/m. The van't Hoff factor (i) is the number of moles of particles produced when 1 mole of the solute dissolves. Na2SO4(s) ===> 2Na2+(aq) + SO4 2-(aq) . .i = 3 (2 moles of Na+ and 1 mole of SO4 2- are produced) AlCl3(s) ===> Al3+(aq) + 3Cl-(aq) ..i = 4 (1 mole of Al3+ and 3 moles of Cl- are produced). Since the Na2SO4 solution and the AlCl3 solution have the same molality (ca.1), then AlCl3 will show the largest freezing point depression since it's i factor is larger. For VP lowering, delta P = (i)(Po)(mole fraction of solute) The solution with the higher i (AlCl3) will have the largest VP lowering; i.e., the Na2SO4 solution will have a higher VP. For osmotic pressure, OP = i x Molarity of solute x RT Higher i = higher osmotic pressure.

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