Q100P - Chemistry: Molecular Nature Of Matter And Change

Question

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 $\left(Kb of ethanol=1.22{}^{°}C/m\right)$?

Step-by-Step Solution

Answer

The boiling point of a solution of 6.4 g of vanillin in 50.0 g of ethanol is 79.528${}^{°}C$

Definition

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.

Calculate molality of vanillin

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

Calculate boiling point elevation ΔT b

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.

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

Calculate actual boiling point

Hence, the actual boiling point will be:

 Boiling point is

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