4.Consider the following solutions:

Question

Consider the following solutions: 

0.010m Na3​PO4​(aq) 

0.020m CaBr2​(aq) 

0.020m KCl(aq) 

0.020m HF(aq) (HF is a weak acid.)

a. Which solution(s) would have the same boiling point as 0.040m C6​H12​O6​(aq)? (C6​H12​O6​ is a nonelectrolyte.)

b. Which solution would have the highest vapor pressure at 28∘C ?

c. Which solution would have the lowest freezing point?

Answer 1

a). The boiling point (a colligative property) is affected (raised) by a dissolved solute, and the extent of the increase in boiling point is related to the number of particles present and the molality (m). The formula for calculating this is ∆T = imK, where ∆T is the change in boiling point, i is the number of particles (van't Hoff factor), m is the molality and K is the boiling constant.

C6H12O6 being a nonelectrolyte provides only ONE particle. The molality is 0.04m, so 0.04 x 1 = 0.04. We need to find a solution that has a molality x particle # = 0.04 from the list. The answer would be 0.01 m Na3PO4 and 0.02 m KCl.

0.01 m Na3PO4 provides that solution. Na3PO4 produces 4 particles (3 Na+ and 1 PO4-) and since it is 0.01 m we have 0.01 x 4 = 0.04.

0.02 m CaBr2 doesn't work. CaBr2 provides 3 particles (1Ca2+ and 2 Br-). 0.02 m x 3 = 0.06.

0.02 m KCl provides that solution. KCl provides 2 particles (1K+ and 1 Cl-). 0.02 m x 2 = 0.04

0.020 m HF doesn't work. As a weak electrolyte it provides only 1 particle. 0.02 x 1 = 0.02

b). Vapor pressure is lowered by the presence of a solute (a colligative property). The lowering of vapor pressure depends on the number of particles and the moles of substance. The solution with the HIGHEST vapor pressure will be the one with the fewest particles and lowest concentration. From the calculations in (a) above, can you decide which that might be?

c). As with boiling point, freezing point is a colligative property and will depend on the number of particles present. The more particles x molality, the lower will be the freezing point. So, from the list and as calculated in (a) above, might this be CaBr2 with a product of m x i = 0.06?