91Ó°ÊÓ

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.

${\mathrm{Δ°Õ}}_{\mathrm{f}}$- The freezing point depression

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

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

$i$-Van't Hoff factor

${\mathrm{Δ°Õ}}_{\mathrm{f}}={\mathrm{ik}}_{\mathrm{f}}\mathrm{m}$

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

role="math" localid="1663307270501" $$\begin{gathered} \mathrm{Molality}\left(\mathrm{m}\right) \mathrm{of} \mathrm{the} \mathrm{solution}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}\left(\mathrm{mol}\right)}{\mathrm{mass} \mathrm{of} \mathrm{solvent}\left(\mathrm{kg}\right)} \\ \mathrm{Moles} \mathrm{of} \mathrm{solute}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}} \end{gathered}$$

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.

Given information:

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

Mass of solution $=100g$

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

role="math" localid="1663307605042" $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathrm{Mass} \%=\frac{\mathrm{mass}  \mathrm{of} \mathrm{solute} }{\mathrm{mass} \mathrm{of} \mathrm{solution}}×100 \\                          0.5=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{100}×100 \\ \mathrm{mass} \mathrm{of} \mathrm{solute}=0.5\mathrm{g}  \mathrm{KCl}. \end{gathered}$$

role="math" localid="1663307616176" $$\begin{gathered} \mathrm{mass} \mathrm{of} \mathrm{solvent}=100-.5 \\                                      =99.5 \mathrm{g}  {\mathrm{H}}_2\mathrm{O}. \end{gathered}$$

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

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

role="math" localid="1663307827484" $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathrm{Δ°Õ}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)+{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ {\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right)={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-\mathrm{Δ}{\mathrm{T}}_{\mathrm{f}} \\                             =0-(-0.234) \\                             =.{234}^{\mathrm{o}}\mathrm{c}. \end{gathered}$$

${\mathrm{Δ°Õ}}_{\mathrm{f}}=0.234.0^{\mathrm{o}}\mathrm{c}$

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

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

${\mathrm{Δ°Õ}}_{\mathrm{f}}={\mathrm{ik}}_{\mathrm{f}}\mathrm{m}$

$$\begin{gathered} \mathrm{i}=\frac{{\mathrm{Δ°Õ}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}} \\ =\frac{0.234}{1.86×0.0674} \\ =\frac{0.234}{0.125364} \\ =1.867 \end{gathered}$$

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

Given information:

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

Mass of solution$=100g$

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

role="math" localid="1663308604673" $$\begin{gathered} \mathrm{Mass} \%=\frac{\mathrm{mass}  \mathrm{of} \mathrm{solute} }{\mathrm{mass} \mathrm{of} \mathrm{solution}}×100 \\                                1=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{100}×100 \\ \mathrm{mass} \mathrm{of} \mathrm{solute}=1\mathrm{g} {\mathrm{H}}_2{\mathrm{SO}}_4. \end{gathered}$$

role="math" localid="1663308576996" $$\begin{gathered} \mathrm{mass} \mathrm{of} \mathrm{solvent}=100-1 \\ =99\mathrm{g} {\mathrm{H}}_2\mathrm{O} \end{gathered}$$

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

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

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

${\mathrm{Δ°Õ}}_{\mathrm{f}}=0.{423}^{\mathrm{o}}\mathrm{c}$

role="math" localid="1663308830829" $$\begin{gathered} \mathrm{Molality}\left(\mathrm{m}\right) \mathrm{of} \mathrm{the} \mathrm{solution}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}\left(\mathrm{mol}\right)}{\mathrm{mass} \mathrm{of} \mathrm{solvent}\left(\mathrm{kg}\right)} \\                          \mathrm{Moles} \mathrm{of} \mathrm{solute}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{solute}} \end{gathered}$$

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

${\mathrm{Δ°Õ}}_{\mathrm{f}}={\mathrm{ik}}_{\mathrm{f}}\mathrm{m}$

$$\begin{gathered} \mathrm{i}=\frac{{\mathrm{Δ°Õ}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}} \\ =\frac{0.423}{1.86×0.103} \\ =\frac{0.423}{0.19158} \\ =2.207 \\ =2.21 \end{gathered}$$

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.