# What is the new boiling point for an aqueous "0.743 molal" solution of "KCl"? K_b = 0.512^@ "C/m"

##### 1 Answer
Aug 9, 2017

As you should have seen in your book,

$\Delta {T}_{b} \equiv {T}_{b} - {T}_{b}^{\text{*}} = i {K}_{b} m$,

where:

• $\Delta {T}_{b}$ is the change in boiling point in $\text{^@ "C}$, from that of the pure solvent, ${T}_{b}^{\text{*}}$, to that of the solution, ${T}_{b}$.
• $i$ is the van't Hoff factor, i.e. the effective number of solute particles in solution.
• ${K}_{b} = {0.512}^{\circ} \text{C/m}$ is the boiling point elevation constant of water.
• $m$ is the molality of the solution... $\text{mol solute/kg solvent}$. Is the solute volatile or nonvolatile?

Assuming 100% dissociation...

${\text{KCl"(aq) -> "K"^(+)(aq) + "Cl}}^{-} \left(a q\right)$

and $1 + 1 = 2 \approx i$, so...

$\textcolor{b l u e}{{T}_{b}} = {T}_{b}^{\text{*}} + i {K}_{b} m$

$= {100}^{\circ} \text{C" + 2 cdot 0.512^@ "C/m" cdot "0.743 m}$

$= \textcolor{b l u e}{\underline{{100.761}^{\circ} \text{C}}}$

What was the change in boiling point?