Question

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

Answer 1

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_"b"` is the molal boiling point elevation constant
• `b` is the molality of the solution

In this problem,

`i = 3`, because 1 mol of `"CaCl"_2` gives 3 mol of ions in solution.
`K_text(b) = "1.86 °C·mol·kg"^"-1"`
`b = "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_"b" = T_"b"^@ + ΔT_"b" = "100 °C + 8.705 °C" = "108.705 °C"`