Q105P - Chemistry: Molecular Nature Of Matter And Change

Question

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

(a) 0.500 mass % KCl, freezing point -0.234^oc 

(b) 1.00 mass % H₂SO₄, freezing point -0.423^oc 

Step-by-Step Solution

Answer

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

(a) 0.0674m KCl,1.867.

(b) 0.103m ,${\mathrm{H}}_2{\mathrm{SO}}_4$ 2.21.

 

Definition

 

The temperature at which the vapour 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.

 

${\mathrm{\Delta T}}_{\mathrm{f}}$- The freezing point depression

${\mathrm{k}}_{\mathrm{f}}$ -  Molal freezing point depression consistent

$\mathrm{m}$-  is the solution molality

$i$- Van't Hoff factor 

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

The total moles of a solute per kilogram of a solvent are 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 potassium chloride solution

Given information:

The freezing point depression $\left({\mathrm{\Delta T}}_{\mathrm{f}}\right)=-0.{234}^{\mathrm{o}}\mathrm{C}$

                              Mass of solution $=100 g$

                Mass $\%$ of KCl in solution $=0.500\%$

$$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ } \mathrm{Mass} \%=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute} }{\mathrm{mass} \mathrm{of} \mathrm{solution}}\times 100 \\ 0.5=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{100}\times 100 \\ \mathrm{mass} \mathrm{of} \mathrm{solute}=0.5 \mathrm{g} \mathrm{KCl}. \end{gathered}$$

$$\begin{gathered} \mathrm{mass} \mathrm{of} \mathrm{solvent} =100-.5 \\ =99.5 \mathrm{g} {\mathrm{H}}_2\mathrm{O}. \end{gathered}$$

Molar mass of potassium chloride $=74.55\mathrm{g}/\mathrm{mol}$

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

$$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ } {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)+{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ {\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right)={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}} \\ =0-(-0.234) \\ =.{234}^{\mathrm{o}}\mathrm{c}. \end{gathered}$$

${\mathrm{\Delta T}}_{\mathrm{f}}=0.234.0^{\mathrm{o}}\mathrm{c}$

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

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}}\times \frac{1000}{\mathrm{mass} \mathrm{of} \mathrm{solvent}} \\ =\frac{0.5}{74.55}\times \frac{1000}{99.5} \\ =\frac{500}{7,417.725} \\ =0.0674\mathrm{m} \mathrm{KCl} \end{gathered}$$

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

$$\begin{gathered} \mathrm{i}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}} \\ =\frac{0.234}{1.86\times 0.0674} \\ =\frac{0.234}{0.125364} \\ =1.867 \end{gathered}$$

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

For sulphuric acid solution

Given information:

The freezing point depression $\left({\mathrm{\Delta T}}_{\mathrm{f}}\right)=-0.{423}^{\mathrm{o}}\mathrm{C}$

Mass of solution $=100 g$

Mass $\%$ of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in solution $=1.00\%$

$$\begin{gathered} \mathrm{Mass} \%=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute} }{\mathrm{mass} \mathrm{of} \mathrm{solution}}\times 100 \\ 1=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{100}\times 100 \\ \mathrm{mass} \mathrm{of} \mathrm{solute}=1\mathrm{g} {\mathrm{H}}_2{\mathrm{SO}}_4. \end{gathered}$$

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

Molar mass of sulphuric acid $=98 g/mol$

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

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)+{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ {\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right)={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}} \\ =0-(-0.423) \\ =.{423}^{\mathrm{o}}\mathrm{c}. \end{gathered}$$

${\mathrm{\Delta T}}_{\mathrm{f}}=0.{423}^{\mathrm{o}}\mathrm{c}$

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

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}}\times \frac{1000}{\mathrm{mass} \mathrm{of} \mathrm{solvent}} \\ =\frac{1}{98}\times \frac{1000}{99} \\ =\frac{1000}{9,702} \\ =0.103\mathrm{m} {\mathrm{H}}_2{\mathrm{SO}}_{4_{}} \end{gathered}$$

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

$$\begin{gathered} \mathrm{i}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}} \\ =\frac{0.423}{1.86\times 0.103} \\ =\frac{0.423}{0.19158} \\ =2.207 \\ =2.21 \end{gathered}$$

Sulphuric acid is a strong acid and dissociates to give a hydrogen ion and a hydrogen sulphate ion. The hydrogen sulphate ion may further dissociate to another hydrogen ion and a sulphate ion.