Question

Can you help me out please? Thanks!

When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. If 2.5 mol of aluminum nitrate is added to the same amount of water, by how much will the boiling point be changed? Show all calculations leading to your answer OR use 3 – 4 sentences to explain your answer.

Answer 1

The boiling point will be raised by 4 °C.

Explanation:

The formula for boiling point elevation `ΔT_text(b)` is

`color(blue)(bar(ul(|color(white)(a/a)ΔT_text(b) = iK_text(b)bcolor(white)(a/a)|)))" "`

where

`i` is the van't Hoff `i`-factor for the solute
`K_text(b)` is the molal boiling point elevation constant of the solvent
`b` is the molal concentration (molality) of the solute

If we have two different solutions of the same molality in the same solvent, we can write

`(ΔT_text(b,2))/(ΔT_text(b,1)) = (i_2color(red)(cancel(color(black)(K_text(b)b))))/(i_1color(red)(cancel(color(black)(K_text(b)b)))) = i_2/i_1`

or

`ΔT_text(b,2) = ΔT_text(b,1) × i_2/i_1`

Let `"component 1 = sucrose"` and `"component 2 = aluminium nitrate"`.

Sucrose is a covalent compound. It does not dissociate in solution, so `i = 1`.

Aluminium nitrate is a soluble ionic compound. It dissociates in solution according to the equation

`"Al"("NO"_3)_3"(s)" → "Al"^"3+""(aq)" + 3"NO"_3^"-""(aq)"`

Thus, 1 mol of solute forms 4 mol of particles, and `i = 4`.

∴ `ΔT_text(b,2) = "1 °C" × 4/1 = "4 °C"`