Question

Which has the highest boiling point?
A. $0.1\,{\text{M}}\,\,{\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$
B. $0.1\,{\text{M}}\,\,{{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}$
C. $0.1\,{\text{M}}\,\,{\text{MgC}}{{\text{l}}_2}$
D. $0.1\,{\text{M}}\,\,\,{\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$

Answer 1

Hint: The elevation in boiling point is the product of the boiling point elevation constant and molality.
The formula to calculate the elevation in boiling point is given as: $\Delta {T_b}\, = i\,{K_b}.\,m$

Complete step by step answer:
The formula of boiling point elevation is as follows:
$\Delta {T_b}\, = i\,{K_b}.\,m$
Where,
$\Delta {T_b}\,$ is the elevation in boiling point.
${K_b}$ is the boiling point elevation constant.
$i\,$ is the van't Hoff factor.
The molality of all the solutions is the same, so the increase in boiling point depends upon the van't Hoff factor.
The van't Hoff factor is directly proportional to the elevation in boiling point. So, as the number of ions in solution increases the boiling point increases.
Dissociation of sodium sulphate ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$ in water is shown as follows:
${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}\mathop \to \limits^{{{\text{H}}_{\text{2}}}{\text{O}}} \,2\,{\text{N}}{{\text{a}}^{\text{ + }}}\, + \,{\text{SO}}_4^ - $
Sodium sulphate ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$ produces three ions so, the value of van’t Hoff factor is 3.
Dissociation of glucose ${{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}$ in water is shown as follows:
Glucose is not an ionic compound, so it does not dissociate in water.
So, the value of van’t Hoff factor for glucose ${{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}$ is 1.
Dissociation of magnesium chloride ${\text{MgC}}{{\text{l}}_2}$ in water is shown as follows:
${\text{MgC}}{{\text{l}}_2}\mathop \to \limits^{{{\text{H}}_{\text{2}}}{\text{O}}} \,{\text{M}}{{\text{g}}^{{\text{2 + }}}}\, + \,2\,{\text{C}}{{\text{l}}^ - }$
Magnesium chloride ${\text{MgC}}{{\text{l}}_2}$ produces three ions so the value of van’t Hoff factor is 3.
Dissociation of aluminium nitrate ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ in water is shown as follows:
${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}\mathop \to \limits^{{{\text{H}}_{\text{2}}}{\text{O}}} \,{\text{A}}{{\text{l}}^{{\text{3 + }}}}\, + \,3\,{\text{NO}}_3^ - $
Aluminium nitrate ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ produces four ions so, the value of van’t Hoff factor is 4.
So, Aluminium nitrate ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ has the highest number of ions so, it has the highest boiling point.

Therefore, option (D) ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ is correct.

Note: When a solute is added to the pure solvent, the boiling point of the solution increases which is known as the elevation in the boiling point. The van't Hoff factor represents the degree of dissociation or number of ions produced by a compound on dissolution.