# What is the boiling point of an aqueous solution of "CaCl"_2 with a molal concentration of 1.56 mol/kg?

Apr 21, 2017

The boiling point of the solution is 108.705 °C.

#### Explanation:

The formula for boiling point elevation ΔT_"b" is

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

where

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

In this problem,

$i = 3$, because 1 mol of ${\text{CaCl}}_{2}$ gives 3 mol of ions in solution.
${K}_{\textrm{b}} = \text{1.86 °C·mol·kg"^"-1}$
$b = \text{1.56 mol·kg"^"-1}$

ΔT_"b" = iK_"b"b = 3 × "1.86 °C·"color(red)(cancel(color(black)("kg·mol"^"-1"))) × 1.56 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "8.705 °C"

${T}_{\text{b" = T_"b"^@ + ΔT_"b" = "100 °C + 8.705 °C" = "108.705 °C}}$