# How many grams of salt must be added to 300g of water to increase its boiling point by 5°C?

Dec 22, 2016

You must add 86 g of salt.

#### Explanation:

The formula for boiling point elevation is

color(blue)(|bar(ul(ΔT_"b" = iK_"b"m)|)

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" = "5 °C"
$i = 2$, because 1 mol of $\text{NaCl}$ gives 2 mol of particles in solution.
${K}_{\text{b" = "0.512 °C·kg·mol"^"-1}}$

m = (5 color(red)(cancel(color(black)("°C"))))/(2 × 0.512 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "4.88 mol·kg"^"-1"

Next, we calculate the moles of $\text{NaCl}$.

The formula for molality is

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

$\text{Moles of NaCl" = 0.300 color(red)(cancel(color(black)("kg"))) × "4.88 mol"·color(red)(cancel(color(black)("kg"^"-1"))) = "1.46 mol}$

Finally, convert moles of $\text{NaCl}$ to grams of $\text{NaCl}$.

$\text{Mass of NaCl" = 1.46 color(red)(cancel(color(black)("mol NaCl"))) × "58.44 g NaCl"/(1 color(red)(cancel(color(black)("mol NaCl")))) = "86 g NaCl}$