Question

Calculate the molality and van’t Hoff factor (i) for the following aqueous solutions:

(a) 1.00 mass % NaCl, freezing point -0.593^oc

(b) 0.500 mass % CH₃COOH, freezing point -0.159^oc

Short answer

The molalities and the van't Hoff factors of the given aqueous solutions are:

(a) 0.173m NaCl,1.84.

(b) 0.084m CH₃COOH,1.02.

Step-by-step solution

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.

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}-Thefreezingpointdepression \\ {K}_f-Molal-Thefreezingpointdepressionconsistent \\ m-Molalityofthesolution \\ i-VantHofffactor \\ {\mathrm{\Delta T}}_{\mathrm{f}}=i{k}_{f}m \end{gathered}$$

The total moles of a solute per kilogram of a solvent is termed as Molality.

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

Van't Hoff factor (i) gives a ratio of the concentrations of particles or ions formed when a solute or a substance is dissolved in a solution. For a non-electrolyte solution, the Van’t Hoff is equal to 1.

For sodium chloride solution

Given information:

The freezing point depression $\left({\mathrm{\Delta T}}_{\mathrm{f}}\right)=-0.{593}^{\mathrm{o}}\mathrm{c}$

Mass of solution $=100g$

Mass $\%$of NaCl in solution $=1\%$

$$\begin{gathered} \mathrm{Mass} \%=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute} }{\mathrm{mass} \mathrm{of} \mathrm{solution}}\times 100 \\ 1=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{100}\times 100 \\ \mathrm{mass} \mathrm{of} \mathrm{solute}=1\mathrm{g} \mathrm{NaCl} \end{gathered}$$

$$\begin{gathered} \mathrm{mass} \mathrm{of} \mathrm{solvent}=100-1 \\ =99\mathrm{g} {\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Molar mass of sodium chloride$=58.5g/mol$

$k_f$for water $=1.{86}^{\mathrm{o}}\mathrm{cmol}-1$.

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)+{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ {\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right)={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}} \\ =0-(-0.593) \\ =.{593}^{\mathrm{o}}\mathrm{c} \end{gathered}$$

${\mathrm{\Delta T}}_{\mathrm{f}}=0.593.0^{\mathrm{o}}\mathrm{c}$

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

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}}\times \frac{1000}{\mathrm{mass} \mathrm{of} \mathrm{solvent}} \\ =\frac{1}{58.5}\times \frac{1000}{99} \\ =\frac{1000}{5,791.5} \\ =0.173\mathrm{m} \mathrm{NaCl}. \end{gathered}$$

${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{ik}}_{\mathrm{f}}\mathrm{m}$

$$\begin{gathered} \mathrm{i}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}} \\ =\frac{0.593}{1.86\times 0.173} \\ =\frac{0.593}{0.32178} \\ =1.843. \end{gathered}$$

Hence, the value of i is close to two as the NaCl dissociates into two particles when dissolving in water.

For acetic acid solution

Given information:

The freezing point depression (${\mathrm{\Delta T}}_{\mathrm{f}}$) =-0.159${}^{°}C$

Mass of solution$=100g$

Mass % of ${\mathrm{CH}}_3\mathrm{COOH}$in solution $=0500\%$

role="math" localid="1663312720406" $$\begin{gathered} \mathrm{Mass} \%=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute} }{\mathrm{mass} \mathrm{of} \mathrm{solution}}\times 100 \\ 5=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{100}\times 100 \\ \mathrm{mass} \mathrm{of} \mathrm{solute}=.5\mathrm{g} {\mathrm{CH}}_3\mathrm{COOH} \\ \mathrm{mass} of \mathrm{solvent}=100-.5 \\ =99.5\mathrm{g} {\mathrm{H}}_2\mathrm{O} \\ Molarmassofaceticacid=60g/mol \\ k_{f}forwater=1.86°Cmo{l}^{-1} \\ {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)+{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ {\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right)={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}} \\ =0-(-0.159) \\ =.{159}^{\mathrm{o}}\mathrm{c}. \\ {\mathrm{\Delta T}}_{\mathrm{f}}=0.159.0^{\mathrm{o}}\mathrm{c}. \end{gathered}$$

$$\begin{gathered} \mathrm{Molality}\left(\mathrm{m}\right) \mathrm{of} \mathrm{the} \mathrm{solution}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}\left(\mathrm{mol}\right)}{\mathrm{mass} \mathrm{of} \mathrm{solvent}\left(\mathrm{kg}\right)} \\ \mathrm{Moles} \mathrm{of} \mathrm{solute}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}} \\ \mathrm{m}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}}\times \frac{1000}{\mathrm{mass} \mathrm{of} \mathrm{solvent}} \\ =\frac{.5}{60}\times \frac{1000}{99.5} \\ =\frac{500}{5,970} \\ =0.0837 \\ =0.084\mathrm{m} {\mathrm{CH}}_3\mathrm{COOH}. \\ {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{ik}}_{\mathrm{f}}\mathrm{m} \\ \mathrm{i}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}} \\ =\frac{0.159}{1.86\times 0.084} \\ =\frac{0.159}{0.156} \\ =1.019 \\ =1.02. \end{gathered}$$

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{ik}}_{\mathrm{f}}\mathrm{m} \\ \mathrm{i}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}} \\ =\frac{0.159}{1.86\times 0.084} \\ =\frac{0.159}{0.156} \\ =1.019 \\ =1.02. \end{gathered}$$

As acetic acid is a weak acid and dissociates to a small extent in a solution, a van’t Hoff factor is close to 1.