Question

Calculate the depression in the freezing point of water when $10g$ of $C{H_3}C{H_2}CHClCOOH$ is added to $250g$ of water.
[\[{K_a} = 1.4 \times {10^{ - 3}},Kf = 1.86kg\,\,mo{l^{ - 1}}\]]

Answer 1

Hint:
When a non-volatile solute is added in a solvent, its freezing point decreases from the actual value, this phenomenon is known as freezing point depression. This is because the chemical potential of the solvent in pure solvent is higher than in the mixture solvent.

Complete step by step answer:
To calculate the freezing point of water when $10g$ of $C{H_3}C{H_2}CHClCOOH$ is added to $250g$ of water, we need the formula:
\[\Delta TfO = i\,k\,f\,m\]
Given, \[{K_a} = 1.4 \times {10^{ - 3}}\]
\[Kf = 1.86kg\,\,mo{l^{ - 1}}\]
Weight of solute\[ = 10g\]
Weight of solvent\[ = 250g\]
Therefore, molecular weight of the solute\[ \Rightarrow C{H_3}C{H_2}CHClCOOH\]
\[ = 4\left( {12} \right) + 7\left( 1 \right) + 2\left( {16} \right) + 35.5\]
\[ = 48 + 7 + 32 + 35.5\]
\[ = 122.5g/mol\]
According to the formula given above, we know the value of \[Kf\], but we need to find the value of \[i\] and \[m\].
Calculation of \[m\] i.e. molality:
\[{\text{Molality = }}\dfrac{{{\text{Weight of solute}} \times {\text{1000}}}}{{{\text{Molecular weight of solute}} \times {\text{weight of solvent}}}}\]
Therefore,
\[{\text{m = }}\dfrac{{10 \times {\text{1000}}}}{{122.5 \times 250}}\]
\[{\text{m = 0}}{\text{.3265}}mol\,k{g^{ - 1}}\]
Calculation of \[i\] i.e. Van’t Hoff factor:
Let \[\alpha \, = \,{\text{Degree of dissociation of the acid}}\]
\[C{H_3}C{H_2}CHClCOOH + {H_2}O \rightleftharpoons C{H_3}C{H_2}CHClCO{O^ - } + {H_2}{O^ \oplus }\]

Initial concentration:	\[C\]	\[O\] \[ + \] \[O\]
Concentration at equilibrium:	\[C - C\alpha \]	\[C\alpha \] \[ + \] \[C\alpha \]

We know, \[K\alpha = \dfrac{{{\text{Concentration of products}}}}{{{\text{Concentration of reactants}}}}\]
\[K\alpha = \dfrac{{\left[ {C\alpha } \right]\left[ {C\alpha } \right]}}{{C - C\alpha }}\]
As, \[\alpha \] is very small with respect to \[1\], \[1 - \alpha = 1\]
\[K\alpha = C{\alpha ^2}\]
\[\alpha = \sqrt {\dfrac{{K\alpha }}{C}} \]
On putting the values,
\[\alpha = \sqrt {\dfrac{{1.4 \times {{10}^{ - 3}}}}{{0.3265}}} \]
\[\alpha = 0.065\]
At equilibrium,
\[C{H_3}C{H_2}CHClOH \rightleftharpoons C{H_3}C{H_2}CHClCO{O^ - } + {H^ \oplus }\]

Initial moles:	\[1\]	\[O\] \[ + \] \[O\]
Final moles:	\[1 - \alpha \]	\[\alpha \] \[ + \] \[\alpha \]

Therefore, total moles at equilibrium
\[ = 1 - \alpha + \alpha + \alpha \]
\[ = 1 + \alpha \]
Therefore, Van’t Hoff factor, \[i\]
\[i = \dfrac{{1 + \alpha }}{1}\]
\[1 + 0.065 = 1.065\]
Hence, depression in freezing point is \[\Delta Tf\]
\[\Delta Tf = i\,k\,f\,m\]
\[ = 1.065 \times 1.86 \times 0.3265\]
\[\Delta Tf = {0.647^\circ }C\]

Note:The lowering or depression in freezing point is because when a non-volatile solute adds to a volatile liquid solvent, then the vapour pressure of the solution becomes lower than that of the original or pure solution. So, the solids reach at an equilibrium with the solution at lower temperature.