Arrange the following in aqueous solutions in order of increasing freezing points lowest to highest temperature 10 m glucose 0.10 m BaCl2 0.20 m NaCl and 0.20 m Na2SO4?

Answer 1

The Formula used to find the freezing point is "Delta T=Km"

"Delta T" is the change of temperature and what your are looking for.

"K" is the freezing constant of the solvent

"m" is the molality which is mols of solute over kg of solvent.

Based on the formula if you have 0.10m of each solvent you would think that it is the same. However, you also need to consider the number of mols of substance once the salt has disolved. NaCl will disassociate and then you will have more mols to account for in your molality calculation. This will give salt the lower freezing point.

Answer 2

If all of the "m"'s immediately preceded by a number in the question are intended to mean "molar" and 10 molar glucose actually exists, then the freezing points from lowest to highest are 10 molar glucose, 0.20 molar Na2SO4, 0.20 molar NaCl, and 0.10 molar BaCl2. Each mole of Na2SO4 provides 3 moles of ions after dissociation, each mole of NaCl provides 2 moles of ions after dissociation, and each mole of BaCl2 provides 3 moles of ions after dissociation. (Glucose does not ionize.) 10 > (0.20 X 3) > (0.20 X 2) > (0.10 X 3), and the largest concentration of dissolved units produces the greatest freezing point depression.

Answer 3

The glucose will C6H12O6 it has more particles when dissolved. Colligative properties depends only on the number of particles not the nature. Solutions will have a lower vapor pressure, higher boiling points and lower freezing points

Answer 4

0.5 m Na2CO3

1 m NaCI

3 m glycerol

2 m NH4NO3

1.5 m Ca(NO3)2

Answer 5

NaCl

Because NaCl --> Na+ + Cl-

0.1 0.1

0.1 + 0.1 = 0.2

NaCl forms 0.2m solute, therefore it has a lower F.P