91Ó°ÊÓ

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.

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

$$\begin{gathered} Given \inf ormation: \\ Thefreezingpointdepression\left(\mathrm{Δ}{T}_f\right)=-15.0^{∘}C \\          Massoftheglycerolinsolution=? \\               Massofthesolvent=11.0mg=11.0×{10}^{-6}kg \\         Molarmassofglycerol\left(C_3{H}_8{O}_3\right)=92g/mol \\                                                           K_{f}forwater=1.86Kkgmo{l}^{-1} \end{gathered}$$

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

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

$$\begin{gathered} m=\frac{massofsolute}{molarmassofsolute}×\frac{1}{massofsolvent} \\ 8.065=\frac{massofsolute}{92}×\frac{1}{11×{10}^{-6}} \end{gathered}$$

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