Question

Mark the correct relationship between the boiling points of very dilute solutions of $$\mathrm{B}\mathrm{a}\mathrm{C}{\mathrm{l}}_2$$
($${\mathrm{t}}_1$$) and KCl ($${\mathrm{t}}_2$$), having the same molarity.
A. $${\mathrm{t}}_1={\mathrm{t}}_2$$
​B. $${\mathrm{t}}_1>{\mathrm{t}}_2$$
​C. $${\mathrm{t}}_2>{\mathrm{t}}_1$$
​D. $${\mathrm{t}}_2$$ is approximately equal to $${\mathrm{t}}_1$$.

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:
$$\mathrm{i}=\frac{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}/\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{b}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}/\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$$

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:
$$\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\to \mathrm{N}{\mathrm{a}}^{+}+\mathrm{C}{\mathrm{l}}^{-}$$

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 $$\mathrm{N}{\mathrm{a}}^{+}$$ (sodium ions) and $$\mathrm{C}{\mathrm{l}}^{-}$$
(chloride ions) will be released into the solution.

So for NaCl,
$$\mathrm{i}=\frac{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}/\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{b}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}/\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$$
$$\Rightarrow \mathrm{i}=\frac{2}{1}$$
$$\Rightarrow \mathrm{i}=2$$

Barium chloride dissociates as follows:
$$\mathrm{B}\mathrm{a}\mathrm{C}{\mathrm{l}}_2\to \mathrm{B}{\mathrm{a}}^{2+}+2\mathrm{C}{\mathrm{l}}^{-}$$
It dissociates into positive ions called cations and negative ions called anions.

After the dissolution of one mole of $$\mathrm{B}\mathrm{a}\mathrm{C}{\mathrm{l}}_2$$ in water, one mole each of $$\mathrm{B}{\mathrm{a}}^{2+}$$ (barium ions) and two moles of $$\mathrm{C}{\mathrm{l}}^{-}$$ (chloride ions) will be released into the solution.
There would be three moles of ions in the solution.
So, $$\mathrm{i}=\frac{3}{1}$$
$$\Rightarrow \mathrm{i}=3$$

We know that,
For NaCl
$$\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}=\mathrm{i}{\mathrm{k}}_{\mathrm{b}}\mathrm{m}$$
where; $$\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}$$=elevation in boiling point $${\mathrm{k}}_{\mathrm{b}}$$=molal elevation constant
I = van't Hoff factor = 2
So, $$\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}=2{\mathrm{k}}_{\mathrm{b}}\mathrm{m}={\mathrm{t}}_2$$

For barium chloride,
i=3
$$\mathrm{\Delta }{\mathrm{T}}_{\mathrm{b}}=3{\mathrm{k}}_{\mathrm{b}}\mathrm{m}={\mathrm{t}}_1$$
So, $${\mathrm{t}}_1>{\mathrm{t}}_2$$
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.