Question

Calculate the freezing point and boiling point of an antifreeze solution that is 50.0% by mass ethylene glycol (HOCH₂CH₂OH)in water. Ethylene glycol is a nonelectrolyte.

Short answer

The freezing point and boiling point for a solution of ethylene glycol in water is _{$\text{-29.76}°\text{C}$} and_{$\text{108.32}°\text{C}$}

Step-by-step solution

Step 1:

According to the question, 50.0% by mass ethylene glycol is present which refers to a 100 g solution consisting of 50 g of solute.

Firstly, we have to calculate the number of moles of ethylene i.e.,_{$\text{Moles=}\frac{\text{Mass}}{\text{Molarmass}}$}

As the mass of solute is 50g and the molar mass of ethylene glycol is 62g then$\text{No.ofMoles=}\frac{50}{62}\text{=0.80g}$

Mass of solvent = Mass of a solution – Mass of solute

= 100 – 50 = 50 g

Therefore, the Molality of the solution will be _{$\text{Molality=}\frac{\text{Molesofsolute}\times \text{1000}}{\text{Volumeofsolvent}}$}

_{$\text{Molality=}\frac{\text{0.80}\times \text{1000}}{\text{50}}\text{=16m}$}

Freezing point depression

The relation for depression in the freezing point is _{$\Delta {\text{T}}_{\text{f}}{\text{=k}}_{\text{f}}\times \text{m}$}

Here_{$\Delta {\text{T}}_{\text{f}}$} is the difference between the freezing point of pure solvent and the freezing point of solution i.e.,_{$\Delta {\text{T}}_{\text{f}}\text{=}\Delta {\text{T}}_f°\text{ofpuresolvent-}\Delta {\text{T}}_f°\text{ofsolution}$}

_{$\Delta {\text{T}}_{\text{f}}\text{=}0\text{-}\Delta {\text{T}}_f°\text{ofsolution}$}

Therefore, _{$0\text{-}\Delta {\text{T}}_f°{\text{ofsolution=k}}_{\text{f}}\times \text{m}$} here, K_f for water is _{$1.86°\text{C/m}$}and m is molality.

_{$\begin{array}{l}0\text{-}\Delta {\text{T}}_f°\text{ofsolution=1.86}\times \text{16=29.76}°\text{C}\\ \Delta {\text{T}}_f°\text{ofsolution=-29.76}°\text{C}\end{array}$}

Elevation in boiling point

The relation for elevation in boiling point is given _{$\Delta {\text{T}}_{\text{b}}{\text{=k}}_{\text{b}}\times \text{m}\times \text{i}$}

Here, ‘I’ for a nonelectrolyte is 1, K_b is $\text{0.52}°\text{C/m}$for water, whereas _{$\Delta {\text{T}}_{\text{b}}$} is the difference between freezing point of solution and freezing point of pure solvent.

Therefore,

_{$\begin{array}{l}\Delta {\text{T}}_{\text{s}}°\text{-100=0.52}\times \text{16}\times \text{1}\\ \Delta {\text{T}}_{\text{s}}°\text{=8.32+100=108.32}°\text{C}\end{array}$}