Question

Mark the correct relationship between the boiling points of very dilute solutions of \[{\rm{BaC}}{{\rm{l}}_{\rm{2}}}\]
(\[{{\rm{t}}_{\rm{1}}}\]) and KCl (\[{{\rm{t}}_{\rm{2}}}\]), having the same molarity.
A. \[{{\rm{t}}_{\rm{1}}}{\rm{ = }}{{\rm{t}}_{\rm{2}}}\]
​B. \[{{\rm{t}}_{\rm{1}}}{\rm{ > }}{{\rm{t}}_{\rm{2}}}\]
​C. \[{{\rm{t}}_{\rm{2}}}{\rm{ > }}{{\rm{t}}_{\rm{1}}}\]
​D. \[{{\rm{t}}_{\rm{2}}}\] is approximately equal to \[{{\rm{t}}_{\rm{1}}}\].

Answer 1

Hint: Elevation in boiling point happens when a solution possesses a higher boiling point than a pure solvent. When a non-volatile solute is added to a pure solvent, its boiling point increases.

Formula Used:
\[{\rm{i = }}\frac{{{\rm{Total number of moles of particles after association/dissociation}}}}{{{\rm{Number of moles of particles before association/dissociation}}}}\]

Complete Step by Step Solution:
We have to find out the elevation in boiling point for very dilute solutions of sodium chloride and barium chloride for the same molal concentration.

Sodium chloride will dissociate as follows:
\[{\rm{NaCl}} \to {\rm{N}}{{\rm{a}}^{\rm{ + }}}{\rm{ + C}}{{\rm{l}}^{\rm{ - }}}\]

NaCl dissociates into positive ions called cations and negative ions called anions.
After the dissolution of one mole of NaCl in water, one mole each of \[{\rm{N}}{{\rm{a}}^{\rm{ + }}}\] (sodium ions) and \[{\rm{C}}{{\rm{l}}^{\rm{-}}}\]
(chloride ions) will be released into the solution.

So for NaCl,
\[{\rm{i = }}\frac{{{\rm{Total number of moles of particles after association/dissociation}}}}{{{\rm{Number of moles of particles before association/dissociation}}}}\]
\[ \Rightarrow {\rm{i = }}\frac{2}{1}\]
\[ \Rightarrow {\rm{i = 2}}\]

Barium chloride dissociates as follows:
\[{\rm{BaC}}{{\rm{l}}_{\rm{2}}} \to {\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}{\rm{ + 2C}}{{\rm{l}}^{\rm{ - }}}\]
It dissociates into positive ions called cations and negative ions called anions.

After the dissolution of one mole of \[{\rm{BaC}}{{\rm{l}}_{\rm{2}}}\] in water, one mole each of \[{\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}\] (barium ions) and two moles of \[{\rm{C}}{{\rm{l}}^{\rm{-}}}\] (chloride ions) will be released into the solution.
There would be three moles of ions in the solution.
So, \[{\rm{i = }}\frac{3}{1}\]
\[ \Rightarrow {\rm{i = 3}}\]

We know that,
For NaCl
\[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = i}}{{\rm{k}}_{\rm{b}}}{\rm{m}}\]
where; \[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\]=elevation in boiling point \[{{\rm{k}}_{\rm{b}}}\]=molal elevation constant
I = van't Hoff factor = 2
So, \[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = 2}}{{\rm{k}}_{\rm{b}}}{\rm{m = }}{{\rm{t}}_{\rm{2}}}\]

For barium chloride,
i=3
\[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = 3}}{{\rm{k}}_{\rm{b}}}{\rm{m = }}{{\rm{t}}_1}\]
So, \[{{\rm{t}}_{\rm{1}}}{\rm{ > }}{{\rm{t}}_{\rm{2}}}\]
Hence, the boiling point of barium chloride is greater than the boiling point of sodium chloride.
So, option B is correct.

Note: In this question, it is mentioned that a very dilute solution of sodium chloride and barium chloride. These compounds being ionic dissociate in water completely. Barium chloride forms three moles of particles resulting in a van't Hoff factor of 3 and consequently has a higher boiling point.