
The molal elevation constant is the ratio of the elevation in boiling point to:
A. Molarity
B. Molality
C. Mole fraction of solute
D. Mole fraction of solvent
Answer
162.3k+ views
Hint: Molal elevation constant is the increase in boiling point in the dissolution of one mole of a non-volatile solute in 1 kg or 1000g of a solvent. It is also called the ebullioscopic constant.
Formula Used:
\[{\rm{i = }}\frac{{{\rm{Total number of moles of particles after association/dissociation}}}}{{{\rm{Number of moles of particles before association/dissociation}}}}\] where I= van't Hoff factor
Complete Step by Step Solution:
The boiling point is the temperature when the vapour pressure is the same as the atmospheric pressure.
The vapour pressure of a solution is always less than the vapour pressure of a solvent.
This happens because the addition of non-volatile solute to a solvent decreases the number of solvent particles present on the surface of the liquid. As the solvent particles decrease, less amount of these particles gets converted to vapours. So, vapour pressure decreases.
With the decrease in vapour pressure, a high temperature is required for the liquid particles for the boiling to happen. So, the boiling point increases. Thus, the boiling point of the solution is greater than the pure solvent. This only depends on the number of solute particles, not on their nature.
Let \[{{\rm{T}}_{\rm{b}}}^{\rm{^\circ }}\] be the boiling point of pure solvent and
\[{{\rm{T}}_{\rm{b}}}\] is the boiling point of the solution.
The increase in the boiling point
\[\Delta {{\rm{T}}_{\rm{b}}}{\rm{ = }}{{\rm{T}}_{\rm{b}}}{\rm{ - }}{{\rm{T}}_{\rm{b}}}^{\rm{^\circ }}\] is known as the elevation of the boiling point.
For dilute solutions, the elevation of boiling point and the molality of the solute in a solution have a proportional relationship.
So, \[{\rm{\Delta }}{{\rm{T}}_b} \propto {\rm{m}}\]
or \[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = }}{{\rm{K}}_{\rm{b}}}{\rm{m}}\]
Here m is molality.
It is the number of moles of solute which undergoes dissolution in 1kg of solvent.
\[{{\rm{K}}_{\rm{b}}}\] is known as Boiling Point Elevation Constant or Molal Elevation Constant or Ebullioscopic Constant.
Its unit is K kg/mol.
\[{{\rm{K}}_{\rm{b}}}{\rm{ = }}\frac{{{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}}}{{\rm{m}}}\]
So, the molal elevation constant is the ratio of the elevation in boiling point to molality.
So, option B is correct.
Note: While attempting the question, one must mention the mathematical expression of the ebullioscopic constant. It is the ratio of elevation of boiling point to the molality of the solution. Its unit is K kg/mol.
Formula Used:
\[{\rm{i = }}\frac{{{\rm{Total number of moles of particles after association/dissociation}}}}{{{\rm{Number of moles of particles before association/dissociation}}}}\] where I= van't Hoff factor
Complete Step by Step Solution:
The boiling point is the temperature when the vapour pressure is the same as the atmospheric pressure.
The vapour pressure of a solution is always less than the vapour pressure of a solvent.
This happens because the addition of non-volatile solute to a solvent decreases the number of solvent particles present on the surface of the liquid. As the solvent particles decrease, less amount of these particles gets converted to vapours. So, vapour pressure decreases.
With the decrease in vapour pressure, a high temperature is required for the liquid particles for the boiling to happen. So, the boiling point increases. Thus, the boiling point of the solution is greater than the pure solvent. This only depends on the number of solute particles, not on their nature.
Let \[{{\rm{T}}_{\rm{b}}}^{\rm{^\circ }}\] be the boiling point of pure solvent and
\[{{\rm{T}}_{\rm{b}}}\] is the boiling point of the solution.
The increase in the boiling point
\[\Delta {{\rm{T}}_{\rm{b}}}{\rm{ = }}{{\rm{T}}_{\rm{b}}}{\rm{ - }}{{\rm{T}}_{\rm{b}}}^{\rm{^\circ }}\] is known as the elevation of the boiling point.
For dilute solutions, the elevation of boiling point and the molality of the solute in a solution have a proportional relationship.
So, \[{\rm{\Delta }}{{\rm{T}}_b} \propto {\rm{m}}\]
or \[{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = }}{{\rm{K}}_{\rm{b}}}{\rm{m}}\]
Here m is molality.
It is the number of moles of solute which undergoes dissolution in 1kg of solvent.
\[{{\rm{K}}_{\rm{b}}}\] is known as Boiling Point Elevation Constant or Molal Elevation Constant or Ebullioscopic Constant.
Its unit is K kg/mol.
\[{{\rm{K}}_{\rm{b}}}{\rm{ = }}\frac{{{\rm{\Delta }}{{\rm{T}}_{\rm{b}}}}}{{\rm{m}}}\]
So, the molal elevation constant is the ratio of the elevation in boiling point to molality.
So, option B is correct.
Note: While attempting the question, one must mention the mathematical expression of the ebullioscopic constant. It is the ratio of elevation of boiling point to the molality of the solution. Its unit is K kg/mol.
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