Question

Which of the following will have the highest boiling point at \[1\] atm pressure?
(A) ${{0}}{{.1 M NaCl}}$
(B) ${{0}}{{.1 M Sucrose}}$
(C) ${{0}}{{.1 M BaC}}{{{l}}_{{2}}}$
(D) ${{0}}{{.1 M Glucose}}$

Answer 1

Hint: Boiling point is the temperature at which the vapor pressure of the liquid becomes $1$ atm. When molarity is given and we have to find the boiling point then we calculate the elevation in boiling point by which the boiling point will increase.

Complete step by step answer:
We know that when solute is mixed in a solution the boiling point of the solution increases due to increased vapor pressure. The expression to find out elevation in boiling point is
${{\Delta }}{{{T}}_{{b}}}{{ = }}{{{K}}_{{b}}}{{ \times m \times i}}$
${{\Delta }}{{{T}}_{{b}}}{{ = }}$ molal elevation constant
${{m = }}$ molality
${{i = }}$ van’t hoff factor
Now the actual increase in boiling point in solution will be the boiling point of the true solvent plus the elevation in boiling point due to added solute.
${{\Delta }}{{{T}}_{{b}}}_{{{(solution)}}}{{ = }}{{{T}}_{{b}}}^{{o}}{{ + \Delta }}{{{T}}_{{b}}}$
Now by seeing both the equations we see that the larger the value of ${{molality \times van't hoff}}$ increases the more the elevation in boiling point also increases.
When the elevation of the boiling point increases then it will have high boiling point. All the \[4\] options above have equal molality so now we will check the van’t hoff factor. Van’t hoff is the number of ions a compound dissociates.
${{NaCl }} \to {{ N}}{{{a}}^{{ + }}}{{ + C}}{{{l}}^{{ - }}}{{ i = 2}}$
${{BaC}}{{{l}}_{{2}}}{{ }} \to {{ B}}{{{a}}^{{{2 + }}}}{{ + 2C}}{{{l}}^{{ - }}}{{ i = 3}}$
${\text{Sucrose/Glucose }} \to {\text{ no dissociation i = 1}}$
So here we can see that the highest value of van’t hoff factor is of ${{BaC}}{{{l}}_{{2}}}$ so it will have highest value of elevation constant and in continuation will have the highest boiling point among all.

So, the correct answer is Option C.

Note: When we mix a non-volatile solute in a solution due to decreased surface area the vapour pressure of the solution increases as the result of that many colligative properties change e.g. depression in freezing point, elevation in boiling point etc
Here molarity is considered as molality as all have equal number of moles in equal number of solutions so if we take molarity or molality it will come out to be the same. So we mainly focus on the variable van’t hoff factor of the compounds.