# What is the boiling point of a 0.527 m aqueous solution of LiBr?

Jul 13, 2017

$\text{boiling point" = 100.540^"o""C}$

#### Explanation:

We're asked to find the new boiling point of a solution with a molal concentration of aqueous $\text{LiBr}$ of $0.527 m$.

To do this, we can use the equation

$\Delta {T}_{b} = i m {K}_{b}$

where

• $\Delta {T}_{b}$ is the change in boiling point of the solution (not necessarily the actual boiling point!)

• $i$ is the van't Hoff factor of the solute, which for these purposes is essentially the number of dissolved ions per unit of the solute.

Since $\text{LiBr}$ splits into ${\text{Li}}^{+}$ and ${\text{Br}}^{-}$, the van't Hoff factor here would be color(orange)(2.

• $m$ is the molality of the solution, given as color(purple)(0.527m

• ${K}_{b}$ is the molal boiling point elevation constant for the solvent; the solvent here is water ("aqueous")

${K}_{b}$ for water (at ${25}^{\text{o""C}}$) is color(green)(0.512 color(green)(""^"o""C/"m (it may be useful to know this)

We therefore have

• DeltaT_b = ?

• i = color(orange)(2

• m = color(purple)(0.527m

• K_b = color(green)(0.512 $\text{^"o""C/} m$

Plugging these into our equation, we have

DeltaT_b = (color(orange)(2))(color(purple)(0.527)cancel(color(purple)(m)))(color(green)(0.512)(color(green)(""^"o""C"))/(cancel(color(green)(m)))) = color(red)(0.540^"o""C"

This quantity represents by how much the boiling point increases (colligative properties), so to find the new boiling point, we simply add this to the normal boiling point of water (${100}^{\text{o""C}}$):

$\text{boiling point" = 100^"o""C" + color(red)(0.540^"o""C") = color(blue)(100.540^"o""C}$