The boiling point of ethanol $\left({\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\right)$is$\mathbf{78}\mathbf{.}\mathbf{5}{}^{\mathbf{°}}\mathit{C}$. What is the boiling point of a solution of 6.4 g of vanillin$\left(M=152.14g/\mathrm{mol}\right)$in 50.0 g of ethanol$\left(Kbofethanol=1.22{}^{°}C/m\right)$?

We can calculate the boiling point of a substance by determining the change in a boiling point by using ${\mathbf{\Delta T}}_{\mathbf{b}}\mathbf{=}{\mathbf{iK}}_{\mathbf{b}}\mathbf{m}\mathbf{,}$ where the $K_b$is the molal boiling point elevation constant, the m is the molality of the particles in a solution, and the i is the van’t Hoff factor.

$$\begin{gathered} \mathrm{Molesofvanillin}=\frac{6.4\mathrm{gvanillin}}{152.14\mathrm{g}/\mathrm{mol}} \\ =0.042\mathrm{moles} \\ \text{Molality of a vanillin =}\frac{\text{moles of vanillin}}{\text{kg of solvent}\left(\text{mass of solvent ethanol}\right)} \\ =\frac{0.042\mathrm{moles}}{\left(\frac{50\mathrm{g}}{1000\mathrm{kg}}\right)} \\ =0.84\mathrm{m}. \end{gathered}$$

Calculate the boiling point elevation by using ${\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{iK}}_{\mathrm{b}}\mathrm{m}$, where i=1, as vanillin is a non-electrolyte, and cannot dissociate in a solution

m=0.84m as calculated in step-1

$K_b=1.22{}^{°}C/m$as given in the solution.

$$\begin{gathered} ∆T_b=1.22{}^{°}C/m\times 0.84m \\ =1.0284{}^{°}C \end{gathered}$$

Hence, the actual boiling point will be:

Boiling point is

$$\begin{gathered} =78.5{}^{°}C+1.0284{}^{°}C \\ =97.528{}^{°}C \end{gathered}$$