Q102P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the minimum mass of ethylene glycol glycol (C₂H₆O₂) that must be dissolved in 14.5 kg of water to prevent the solution from freezing at -12.0^oF? (Assume ideal behaviour).

Step-by-Step Solution

Answer

The mass of the ethylene glycol is 11.79 kg.

Definition

The temperature at which the vapor pressure of a solution is equal to that of a pure solvent is known as the freezing point of a solution.

 

Concept: For finding the freezing point we will use the equations shown below.

$$\begin{gathered} ∆T_f=k_{f}m \\ ∆T_f-the freezing point of depression \\ {k}_f-the molal drop of the freezing point cons\tan t \\ m- solution molality \end{gathered}$$

Further solution

Given information:

$$\begin{gathered} Thefreezingpointdepression\left(\Delta Tf\right) = -12.0° F \\ Massoftheethyleneglycolinsolution=? \\ Massofthesolvent = 14.5 Kg \\ Molarmassofethyleneglycol\left(C_2{H}_6{O}_2\right) = 62g/mol \\ k_{f}forwater = 1.86 K kg mo{l}^{-1} \end{gathered}$$

$$\begin{gathered} \text{Given:} \\ \text{The freezing point depression }\left({\mathrm{\Delta T}}_{\mathbf{f}}\right) = -12.0°\text{F} \\ \text{Mass of the ethylene glycol in solution }=? \\ \text{Mass of the solvent} = 14.5 \text{Kg} \\ \text{Molar mass of ethyleneglycol}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}\right) = 62.0 \text{g}/mol \\ {\text{K}}_{\text{f}} \text{for water} \text{ = } \text{1.86} \text{K} \text{kg} {\text{mol}}^{\text{ - 1}} \\ {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)+{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ {\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right)={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}} \\ =0-(-24.4) \\ =24.4^{\mathrm{o}}\mathrm{c} \\ {\mathrm{\Delta T}}_{\mathrm{f}}=-12.0^{\mathrm{o}}\mathrm{F} \\ =(-12-32)\times \frac{5}{9} \\ =-44\times \frac{5}{9} \\ =-24.4^{\mathrm{o}}\mathrm{c}. \end{gathered}$$

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{k}}_{\mathrm{f}}\mathrm{m} \\ \mathrm{m}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}} \\ =\frac{24.4}{1.86} \\ =13.12\frac{\mathrm{mol}}{\mathrm{kg}} \\ \mathrm{Molality}\left(\mathrm{m}\right) \mathrm{of} \mathrm{the} \mathrm{solution}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute} \left(\mathrm{mol}\right)}{\mathrm{mass} \mathrm{of} \mathrm{solvent} \left(\mathrm{kg}\right)} \\ \mathrm{Moles} \mathrm{of} \mathrm{solute}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}} \\ \mathrm{m}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}}\times \frac{1}{\mathrm{mass} \mathrm{of} \mathrm{solvent}} \\ 13.12=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{62}\times \frac{1}{14.5} \\ \mathrm{mass} \mathrm{of} \mathrm{solute}=13.12\times 62\times 14.5 \\ =11,794.88 \mathrm{g} \\ =11.79 \mathrm{kg}. \end{gathered}$$