91Ó°ÊÓ

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

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

$$\begin{gathered} ∆T_f=∆k_{f}m \\ ∆T_f-thefreezingpointofdepression \\ {k}_f-themoladropofthefreezingpointcons\tan t \\ m-isthesolutionmolality \end{gathered}$$

Given information:

The freezing point of benzene$=5.5{}^{°}C$

The naphthalene in solution's mass = 5.00g

mass of the solvent of benzene = 444g

= 0.44 kg

$k_f$for benzene$=4.90{}^{°}C/m$

$$\begin{gathered} \mathrm{Molality}\left(\mathrm{m}\right) \mathrm{of} \mathrm{the} \mathrm{solution}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}\left(\mathrm{mol}\right)}{\mathrm{mass} \mathrm{of} \mathrm{solvent}\left(\mathrm{kg}\right)} \\                                                     \mathrm{Moles} \mathrm{of} \mathrm{solute}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}} \\                                                                                  \mathrm{m}=\frac{0.039}{.444} \\                                                                                         =0.087\frac{\mathrm{mol}}{\mathrm{kg}}. \end{gathered}$$

$$\begin{gathered} \mathrm{m}=\frac{0.039}{.444} \\       =0.087\frac{\mathrm{mol}}{\mathrm{kg}} \\ {\mathrm{Δ°Õ}}_{\mathrm{f}}={\mathrm{k}}_{\mathrm{f}}\mathrm{m} \\          =4.90×0.087 \\          =0.426 \\          =0.{43}^{\mathrm{o}}\mathrm{c}. \\                   {\mathrm{Δ°Õ}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)+{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ {\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right)={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-\mathrm{Δ}{\mathrm{T}}_{\mathrm{f}} \\                             =5.5-0.43 \\                             =5.{07}^{\mathrm{o}}\mathrm{c}. \end{gathered}$$