# Question ee6fd

Sep 28, 2015

$4 \text{^@"C}$

#### Explanation:

Before doing any calculation, try to predict what will happen to the boiling poing of the aluminium nitrate solution.

You know that aluminium nitrate, "Al"("NO"_3)_3, is an ionic compound that dissociates completely in aqueous solution to give aluminium cations, ${\text{Al}}^{3 +}$, and nitrate anions, ${\text{NO}}_{3}^{-}$

${\text{Al"("NO"_3)_text(3(s]) -> "Al"_text((aq])^(3+) + 3"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

This tells you that every mole of aluminium nitrate will produce four moles of ions in solution.

Since boiling-point elevation is a colligative property that depends on how many molecules of solute are present in solution, you can expect the boiling point of the aluminium nitrate's solution to be higher than that of the sugar solution.

Now, with that being said, the formula for boiling-point elevation looks like this

$\Delta {T}_{b} = i \cdot {K}_{b} \cdot b \text{ }$, where

$\Delta {T}_{b}$ - the boiling-point elevation;
$i$ - the van't Hoff factor
${K}_{b}$ - the ebullioscopic constant of the solvent;
$b$ - the molality of the solution.

The van't Hoff factor tells you how many particles you get in solution for every molecule or formula unit of solute that dissolves.

Sugar is a non-electrolyte, which means that its van't Hoff factor is equal to $1$. On the other hand, aluminium nitrate's van't Hoff factor is equal to $4$, since you get four ions for every formula unit dissolved.

Therefore, If you add the same number of moles of aluminium nitrate as you add of sugar in the same mass of water, the boiling-point elevation of the aluminium nitrate solution will be four times higher than that of the sugar solution.

${T}_{\text{b aluminium nitrate" = 4 xx T_"b sugar}}$

T_"b aluminium nitrate" = 4 * 1""^@"C" = color(green)(4""^@"C")#

Mathematically, you can prove this by using

${K}_{b} \cdot b = \frac{\Delta {T}_{b 1}}{i} _ 1 \to$ for the sugar solution

For the aluminium nitrate solution, you have the exact same amount of moles of solute and the exact same amount of water, which means that the product ${K}_{b} \cdot b$ is equal to what you have for the sugar solution

$\Delta {T}_{b 2} = {i}_{2} \cdot \frac{\Delta {T}_{{b}_{1}}}{i} _ 1 = {i}_{2} / {i}_{1} \cdot \Delta {T}_{b 1}$

$\Delta {T}_{b 2} = \frac{4}{1} \cdot 1 \text{^@"C" = 4""^@"C}$