Answer
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Hint: The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure. When a non-volatile solute is dissolved in a volatile solvent, the boiling point of the solvent is elevated. The molal elevation constant or the ebullioscopic constant of a solvent is defined as the elevation in boiling point caused by dissolving one mole of a non-volatile solute in $1\,kg$ of the solvent.
Complete answer:
The elevation in boiling point, $\Delta {T_b} = i \times {K_b} \times m$
Where,
$\Delta {T_b}$ is the elevation in boiling point,
$i$ is the van’t-Hoff factor,
${K_b}$ is the molal elevation constant and
$m$ is the molality of the solution.
The van’t Hoff factor is directly proportional to the elevation in boiling point. Hence, we will be considering the van’t Hoff factor of each solution. The solution with the highest $i$ value will have the highest boiling point,
$\Delta {T_b}\,\propto \,i$
a) $1M$ glucose has $i = 1.$ It is an organic solvent.
b) $1M$ KCl has $i = 2.$KCl on dissociation gives two ions: ${K^ + }$ and $C{l^ - }.$
c) $1M\,{\text{Al(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}$ has $i = 4.$ ${\text{Al(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}$ on dissociation gives four ions: $A{l^{3 + }}$ and $3\,NO_{_3}^ - .$
d) $1M\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ has $i = 3.$ ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is a strong electrolyte and on dissociation gives three ions: $2N{a^ + }$ and $SO_{_4}^{2 - }.$
Therefore, the solution with the highest boiling point is c) $1M\,{\text{Al(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}.$
Note:
The van’t Hoff factor is usually used in calculation of degree of association and the degree of dissociation. The degree of association of a solute in a given solvent is defined as the fraction of the total number of molecules or moles of the solute that undergo association. The degree of dissociation of solute is defined as the fraction of the total number of molecules or moles of the solute that undergo dissociation.
Complete answer:
The elevation in boiling point, $\Delta {T_b} = i \times {K_b} \times m$
Where,
$\Delta {T_b}$ is the elevation in boiling point,
$i$ is the van’t-Hoff factor,
${K_b}$ is the molal elevation constant and
$m$ is the molality of the solution.
The van’t Hoff factor is directly proportional to the elevation in boiling point. Hence, we will be considering the van’t Hoff factor of each solution. The solution with the highest $i$ value will have the highest boiling point,
$\Delta {T_b}\,\propto \,i$
a) $1M$ glucose has $i = 1.$ It is an organic solvent.
b) $1M$ KCl has $i = 2.$KCl on dissociation gives two ions: ${K^ + }$ and $C{l^ - }.$
c) $1M\,{\text{Al(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}$ has $i = 4.$ ${\text{Al(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}$ on dissociation gives four ions: $A{l^{3 + }}$ and $3\,NO_{_3}^ - .$
d) $1M\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ has $i = 3.$ ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is a strong electrolyte and on dissociation gives three ions: $2N{a^ + }$ and $SO_{_4}^{2 - }.$
Therefore, the solution with the highest boiling point is c) $1M\,{\text{Al(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}.$
Note:
The van’t Hoff factor is usually used in calculation of degree of association and the degree of dissociation. The degree of association of a solute in a given solvent is defined as the fraction of the total number of molecules or moles of the solute that undergo association. The degree of dissociation of solute is defined as the fraction of the total number of molecules or moles of the solute that undergo dissociation.
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