# If the boiling point increases by "0.126 K" for a solution of a nonelectrolyte in benzene, what molality was required? K_b = "2.52 K" cdot "kg/mol".

Jun 23, 2017

We assume the process is at $\text{1 atm}$, where the normal boiling point would have been ${80.1}^{\circ} \text{C}$. But since we're given $\Delta {T}_{b}$, we don't need to use that.

The boiling point elevation is given by

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

where:

• ${T}_{b}$ is the boiling point.
• $\text{*}$ indicates the pure solvent.
• $i$ is the van't Hoff factor, i.e. the effective number of particles per dissociated formula unit.
• ${K}_{b} = \text{2.52 K/m}$ is the boiling point elevation constant.
• $m$ is the molality in $\text{mol solute/kg solvent}$.

Benzene is the solvent to this nonelectrolyte. So, the molality is given by

$\textcolor{b l u e}{m} = \frac{\Delta {T}_{b}}{i {K}_{b}}$

$\approx \left(0.126 \cancel{\text{K")/(1cdot2.52 cancel"K"cdot"kg/mol}}\right)$

$\approx$ $\textcolor{b l u e}{\text{0.05 mol/kg}}$