Q103P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the minimum mass of glycerol (C₃H₈O₃) that must be dissolved in 11.0 mg of water to prevent the solution from freezing at =-15.0^oc? (Assume ideal behaviour.)

Step-by-Step Solution

Answer

The minimum mass of the glycerol is .0082 g.

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 below.

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{k}}_{\mathrm{f}}\mathrm{m} \\ {\mathrm{\Delta T}}_{\mathrm{f}}-\text{The} \text{freezing} \text{point} \text{depression} \\ {\mathrm{k}}_{\mathrm{f}} - \text{Molal} \text{freezing} \text{point} \text{depression} \text{constant} \\ \text{m - } \text{Solution} \text{molality} \end{gathered}$$

Further solution

$$\begin{gathered} Given \inf ormation: \\ Thefreezingpointdepression\left(\Delta T_f\right)=-15.0^{ \circ } C \\ Massoftheglycerolinsolution=? \\ Massofthesolvent = 11.0 mg = 11.0 \times {10}^{-6} kg \\ Molarmassof glycerol \left(C_3{H}_8{O}_3\right) = 92 g/mol \\ K_f for water = 1.86 K kg mo{l}^{-1} \end{gathered}$$

$$\begin{gathered} {\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right)={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}} \\ =0-(-15) \\ ={15}^{\mathrm{o}}\mathrm{c}. \\ \mathrm{m}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}} \\ =\frac{15}{1.86} \\ =8.065\frac{\mathrm{mol}}{\mathrm{kg}}. \\ {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)+{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ =-15.0^{\mathrm{o}}\mathrm{c}. \end{gathered}$$

$$\begin{gathered} \text{Molality(m)} \text{of} \text{the} \text{solution} \text{ = } \frac{\text{moles} \text{of} \text{solute(mol)}}{\text{mass} \text{of} \text{solvent(kg)}} \\ \text{Moles} \text{of} \text{solute} \text{ = } \frac{\text{mass} \text{of} \text{solute}}{\text{molar} \text{mass} \text{of} \text{solute}} \end{gathered}$$

$$\begin{gathered} m=\frac{mass of solute}{molar mass of solute}\times \frac{1}{mass of solvent} \\ 8.065=\frac{mass of solute}{92}\times \frac{1}{11\times {10}^{-6}} \end{gathered}$$

$$\begin{gathered} \mathrm{mass} \mathrm{of} \mathrm{solute}=8.065\times 92\times 11\times {10}^{-6} \\ =8,161.7811\times {10}^{-6}\mathrm{kg} \\ =.0082 \mathrm{g}. \end{gathered}$$