Question

A 1.00 molal aqueous solution of trichloroacetic acid $$\left( {{\text{CC}}{{\text{l}}_{\text{3}}}{\text{COOH}}} \right)$$ is heated to its boiling point. The solution has the boiling point of $${\text{100}}{\text{.18}}^\circ {\text{C}}$$. Determine the van’t Hoff factor for trichloroacetic acid. $$\left( {{K_{\text{b}}}\;{\text{for}}\;{\text{water}} = 0.512\;{\text{K}}\;{\text{kg}}\;{\text{mo}}{{\text{l}}^{ - 1}}} \right)$$

Answer 1

Hint: In order to calculate van’t Hoff factor, the value of change in freezing point $\left( {\Delta {T_{\text{b}}}} \right)$, molal elevation constant $\left( {{K_{\text{b}}}} \right)$ and molality of the solution (m) are required. In the given question, the value of $\Delta {T_{\text{b}}}$ is not given which can be calculated by the difference between boiling point of the solution and boiling point of the pure solvent.

Formulae used:
$$\Delta {T_{\text{b}}} = {T_{\text{b}}} - T_{\text{b}}^{\text{o}}$$
$\Delta {T_{\text{b}}} = i{K_{\text{b}}}m$

Complete step by step solution:
When the solute undergoes dissociation or association in solution, the number of particles in solution increases or decreases and thus, the colligative property changes accordingly. Van't Hoff factor is used to express the extent of dissociation or association of the solute in solution.
The van’t Hoff factor can be calculated by using the following formula.
$\Delta {T_{\text{b}}} = i{K_{\text{b}}}m \cdot \cdot \cdot \cdot \cdot \cdot \left( {\text{I}} \right)$
Where,
$\Delta {T_{\text{b}}}$ is the change in boiling point.
i is the van’t Hoff factor.
${K_{\text{b}}}$ is the molal elevation constant.
m is the molality of the solution.
According to the question, given data is:
The molality of the aqueous solution is 1.00 m.
The boiling point of the solution is $${\text{100}}{\text{.18}}^\circ {\text{C}}$$.
The value of ${K_{\text{b}}}$for water is $$0.512\;{\text{K}}\;{\text{kg}}\;{\text{mo}}{{\text{l}}^{ - 1}}$$.
Since molality of the solution is defined as number of moles of solute (in mol) present in per kg of the solvent. Thus molality of the aqueous solution can also be written as is $$1\;{\text{mol}}\;{\text{k}}{{\text{g}}^{ - 1}}$$.
In order to calculate van’t Hoff factor, we need to first calculate the change in boiling point that is $\Delta {T_{\text{b}}}$.
The change in boiling point is calculated by using the following expression.
$$\Delta {T_{\text{b}}} = {T_{\text{b}}} - T_{\text{b}}^{\text{o}} \cdot \cdot \cdot \cdot \cdot \cdot \left( {{\text{II}}} \right)$$
Where,
$\Delta {T_{\text{b}}}$ is the change in boiling point.
$${T_{\text{b}}}$$ is the boiling point of the solution.
$$T_{\text{b}}^{\text{o}}$$ is the boiling point of the pure solvent.
The value of $${T_{\text{b}}}$$ is given which is equal to $${\text{100}}{\text{.18}}^\circ {\text{C}}$$. Since we know that pure solvent is water whose boiling point is equal to $$100^\circ {\text{C}}$$. Since the value of ${K_{\text{b}}}$for water is $$0.512\;{\text{K}}\;{\text{kg}}\;{\text{mo}}{{\text{l}}^{ - 1}}$$. Thus we need to convert the unit of $${T_{\text{b}}}$$ and $$T_{\text{b}}^{\text{o}}$$ from degrees Celsius into Kelvin.
The conversion of temperature into Kelvin is done as follows.
$$\displaylines{
  {T_{\text{b}}}\left( {\text{K}} \right) = {T_{\text{b}}}\left( {^\circ C} \right) + 273 \cr
   = {\text{100}}{\text{.18}} + {\text{273}} \cr
   = {\text{373}}{\text{.18}}\;{\text{K}} \cr} $$
Similarly,
$$\displaylines{
  T_{\text{b}}^{\text{o}}\left( {\text{K}} \right) = {T_{\text{b}}}\left( {^\circ C} \right) + 273 \cr
   = {\text{100}} + {\text{273}} \cr
   = {\text{373}}\;{\text{K}} \cr} $$
Thus on substituting known values of $${T_{\text{b}}}$$ and $$T_{\text{b}}^{\text{o}}$$ in equation (II), we get
$$\displaylines{
  \Delta {T_{\text{b}}} = \left( {{\text{373}}{\text{.18}} - {\text{373}}} \right)K \cr
   = 0.18\;{\text{K}} \cr} $$
Now, substitute all the known values of $$\Delta {T_{\text{b}}}$$, ${K_{\text{b}}}$, and molality of the solution in equation (I) to calculate van’t Hoff factor.
$0.18\;{\text{K}} = i \times 0.512\;{\text{K}}\;{\text{kg}}\;{\text{mo}}{{\text{l}}^{ - 1}} \times 1\;{\text{mol}}\;{\text{k}}{{\text{g}}^{ - 1}}$
In order to calculate the value of “i”, divide 0.18 K with $0.512\;{\text{K}}\;{\text{kg}}\;{\text{mo}}{{\text{l}}^{ - 1}} \times 1\;{\text{mol}}\;{\text{k}}{{\text{g}}^{ - 1}}$.

$\displaylines{
  i = \dfrac{{0.18\;{\text{K}}}}{{0.512\;{\text{K}}\;{\text{kg}}\;{\text{mo}}{{\text{l}}^{ - 1}} \times 1\;{\text{mol}}\;{\text{k}}{{\text{g}}^{ - 1}}}} \cr
   = 0.35 \cr} $

Therefore, the value of van’t Hoff factor is equal to 0.35.

Note: In order to calculate change in boiling point $$\left( {\Delta {T_{\text{b}}}} \right)$$, always utilize the formula $$\Delta {T_{\text{b}}} = {T_{\text{b}}} - T_{\text{b}}^{\text{o}}$$ or $$\Delta {T_{\text{b}}} = i{K_{\text{b}}}m$$and to calculate change in freezing point $$\left( {\Delta {T_{\text{f}}}} \right)$$, always utilize the formula $$\Delta {T_{\text{f}}} = T_{\text{f}}^{\text{o}} - {T_{\text{f}}}$$ and $$\Delta {T_{\text{f}}} = i{K_{\text{f}}}m$$.