# What is the molality of an aqueous solution whose new boiling point is 102.5^@ "C" due to adding a nonvolatile solute?

Jan 27, 2016

You know that you have to relate molality with boiling point, so this has to do with colligative properties. I had written an explanation on why:
http://socratic.org/questions/how-do-colligative-properties-affect-freezing-point

The above link explains why adding solute lowers the freezing point and raises the boiling point---the discussed diagram depicts the boiling point as the intersection between the liquid and gas lines, and the intersection moves to the right when the purity of the solvent decreases. i.e. when you add a solute.

The equation you should use is (Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 25-3):

$\setminus m a t h b f \left(\Delta {T}_{\text{vap" = T_"vap" - T_"vap"^"*}} = {K}_{b} \cdot m\right)$

$\setminus m a t h b f \left({K}_{b} = \left({M}_{\text{solvent")/("1000 g/kg")(R(T_"vap"^"*")^2)/(DeltabarH_"vap}}\right)\right)$

where:

• ${T}_{\text{vap"^"*}}$ is the boiling point in $\text{K}$ of the pure solution (no solute)
• ${T}_{\text{vap}}$ is the boiling point in $\text{K}$ of the solution (with solute)
• ${M}_{\text{solvent}}$ is the molar mass of the solvent in $\text{g/mol}$.
• $R$ is the universal gas constant: $\text{8.314472 J/mol"cdot"K}$
• $\Delta {\overline{H}}_{\text{vap}}$ is the molar enthalpy of vaporization, in $\text{kJ/mol}$ (hence, "molar").
• ${K}_{b}$ is known as the boiling point elevation constant.
• $m$ is the molality, which is $\text{mols solute"/"kg solvent}$.

You should get to know the first equation, but generally, you may be given ${K}_{b}$ for lower-level chemistry classes. Also, note that this tends to work better for more dilute solutions.

Since you weren't given ${K}_{b}$, here are the constants you need for water, specifically:

${M}_{\text{solvent" = "18.015 g/mol}}$
${T}_{\text{vap" = 102.5^@ "C" = "375.65 K}}$
${T}_{\text{vap"^"*" = 100.0^@ "C" = "373.15 K}}$
$\Delta {\overline{H}}_{\text{vap" = "40.7 kJ/mol}}$

I would solve for ${K}_{b}$ to get:

${K}_{b} = \left(\text{18.015" cancel"g""/mol")/("1000" cancel"g""/kg") (("8.314472" cancel"J""/"cancel"mol"cdotcancel"K")("373.15 K")^(cancel(2)))/("40700" cancel"J/mol}\right)$

$= \textcolor{g r e e n}{\text{0.5124 K"*"kg/mol}}$

That was the hard part. Now, just solve for $m$. The effect of boiling point elevation is smaller than the effect of freezing point depression though, so I would expect a concentration higher than $\text{1 mol/kg}$. In other words:

The same amount of solute will raise the boiling point less than it will lower the freezing point.

$m = \frac{{T}_{\text{vap" - T_"vap"^"*}}}{K} _ b$

$= \left(2.50 \cancel{\text{K")/("0.5124" cancel"K"*"kg/mol}}\right)$

$=$ $\textcolor{b l u e}{\text{4.879 mol particles/kg water}}$