# How many grams of fructose (C_6H_12O_6) must be dissolved in 882g of 12 acetic acid to raise the boiling point by 7.4°C? The boiling point constant for acetic acid is 3.08 °C/m?

Jan 25, 2017

You must add 380 g of fructose.

#### Explanation:

The formula for boiling point elevation is

color(blue)(bar(ul(|color(white)(a/a)ΔT_"b" = iK_"b"mcolor(white)(a/a)|)))" "

where

• $i$ is the van't Hoff factor
• ΔT_"b" is the boiling point elevation
• ${K}_{\text{b}}$ is the boiling point elevation constant
• $m$ is the molality of the solution

We can solve the equation for the molality of the solution.

m = (ΔT_"b")/(iK_"b")

In this problem,

ΔT_"b" = "7.4 °C"
$i = 1$, because fructose is a nonelectrolyte
${K}_{\text{b" = "3.08 °C·kg·mol"^"-1}}$

m = (7.4 color(red)(cancel(color(black)("°C"))))/(1 × 3.08 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "2.40 mol·kg"^"-1"

Next, we calculate the moles of fructose.

The formula for molality is

color(blue)(bar(ul(|color(white)(a/a) "Molality" = "moles of solute"/"kilograms of solvent"color(white)(a/a)|)))" "

$\text{Moles of fructose" = 0.882 color(red)(cancel(color(black)("kg"))) × "2.40 mol"/(1 color(red)(cancel(color(black)("kg")))) = "2.12 mol}$

Finally, convert moles of fructose to grams of fructose.

$\text{Mass of fructose" = 2.12 color(red)(cancel(color(black)("mol fructose"))) × "180.16 g fructose"/(1 color(red)(cancel(color(black)("mol fructose")))) = "380 g fructose}$