Question

Acetic acid has a boiling point of $ 118.5{\text{ }}{}^ \circ C $ and $ {K_b} $ of $ 3.08{\text{ }}{}^ \circ C{\text{ }}{{\text{m}}^{ - 1}} $ . What is the boiling point of a $ 3.20{\text{ m}} $ solution of $ Ca{\left[ {N{O_3}} \right]_2} $ in acetic acid?

Answer 1

Hint: When $ Ca{\left[ {N{O_3}} \right]_2} $ is added to solution of acetic acid then the boiling of acetic changes due to the dissociation of ions of $ Ca{\left[ {N{O_3}} \right]_2} $ . Since boiling point is a colligative property we can find the change in boiling point of acetic by finding the van’t Hoff factor of $ Ca{\left[ {N{O_3}} \right]_2} $ . Then we can find the new boiling of the acetic acid.
 $ \Delta {T_b}{\text{ }} = {\text{ }}i{{\text{K}}_b}m $
Where,
 $ \Delta {T_b} $ is the change in boiling point
 $ i $ is the Van’t Hoff factor
 $ {{\text{K}}_b} $ is molal elevation constant
 $ m $ is the morality of the solution.

Complete Step By Step Answer:
When the calcium nitrate is added to acetic acid then the boiling point of acetic acid gets changed. This is due to the addition of calcium nitrate. We know that boiling point of solution is a colligative property therefore we can write that:
 $ \Delta {T_b}{\text{ }} = {\text{ }}i{{\text{K}}_b}m $
We can find the value of Van’t Hoff factor, $ i $ as:
 $ Ca{\left[ {N{O_3}} \right]_2}{\text{ }} \to {\text{ C}}{{\text{a}}^{2 + }}{\text{ + 2N}}{{\text{O}}_3}^{ - 1} $
Therefore the total number of ions is three and hence the value of $ i $ is three. We are given that $ {K_b} $ of acetic acid is $ 3.08{\text{ }}{}^ \circ C{\text{ }}{{\text{m}}^{ - 1}} $ and molality of solution is $ 3.20{\text{ m}} $ . Therefore we can write as:
 $ \Rightarrow {\text{ }}\Delta {T_b}{\text{ }} = {\text{ }}i{{\text{K}}_b}m $
 $ \Rightarrow {\text{ }}\Delta {T_b}{\text{ }} = {\text{ 3 }} \times {\text{ }}3.08{\text{ }}{}^ \circ C{\text{ }}{{\text{m}}^{ - 1}}{\text{ }} \times {\text{ 3}}{\text{.20 m }} $
 $ \Rightarrow {\text{ }}\Delta {T_b}{\text{ }} = {\text{ 29}}{\text{.6 }}{}^ \circ C{\text{ }} $
Thus we get the change in the temperature of the boiling point of the solution. Since we are given with the normal boiling point of acetic acid as $ 118.5{\text{ }}{}^ \circ C $ , hence the increased boiling point will be equal to:
 $ \Rightarrow {\text{ }}{T_{new}}{\text{ }} = {\text{ }}\left( {{\text{118}}{\text{.5 + 29}}{\text{.6}}} \right){\text{ }}{}^ \circ C{\text{ }} $
 $ \Rightarrow {\text{ }}{T_{new}}{\text{ }} = {\text{ 148}}{\text{.1 }}{}^ \circ C{\text{ }} $
Hence the new boiling point of the acetic acid will be equal to $ {\text{148}}{\text{.1 }}{}^ \circ C{\text{ }} $ .

Note:
There is no need to convert temperature into kelvin scale as we are calculating the difference of temperature which is equal in both kelvin and degree celsius scale. The value of the van't Hoff factor is equal to the number of ions dissociated. It has no units since it is just a number. The molality of solution must not be confused with the molarity of solution.