Question

What do you understand by elevation of boiling point and molal elevation constant, with the help of this constant how can you find the molecular mass of non-volatile solute?

Answer 1

When a non-volatile solute is added to a solvent, it displaces some solvent molecules at the surface of the solution thereby reducing vapour pressure of the solvent.

Complete step by step solution:
This question deals with colligative properties. Let us first look into colligative properties and the factors on which they depend.
Colligative properties of solutions arise due to the presence of non-volatile solute in a solution. They depend on the number of the solute particles and not on the nature of the chemical species present in a solution. When a non-volatile solute is added to a solvent, it displaces some solvent molecules at the surface of the solution thereby reducing vapour pressure of the solvent which is called the relative lowering of vapour pressure.
A liquid will start to boil at a particular temperature if its vapour pressure becomes equal to the atmospheric pressure at that temperature. Since the vapour pressure of a liquid decreases on the addition of a non-volatile solute, therefore its temperature has to be raised so that more vapours form resulting in an increase in the vapour pressure and when the vapour pressure becomes equal to the atmospheric pressure, the liquid will start to boil. Hence the elevation in boiling point of a liquid is observed on the addition of a non-volatile solute.
We can apply the calculations for colligative properties only for the dilute solutions since their behaviour is very close to that of an ideal solution.
We can calculate the elevation in boiling point of a solution as follows:
Greater the lowering of the vapour pressure of the solution, greater will be the elevation in the boiling point,
$\begin{matrix} \triangle { T }_{ b } \\\ Elevation\quad in\quad boiling\quad point \end{matrix}\propto \begin{matrix} \triangle p \\\ Lowering\quad in\quad vapour\quad pressure \end{matrix}$
Now, according to Roult’s law this lowering in vapour pressure is directly proportional to the mole fraction of the solute:
$\begin{matrix} \triangle p \\\ Lowering\quad in\quad vapour\quad pressure \end{matrix}\propto \begin{matrix} { x }_{ 2 } \\\ Mole\quad fraction\quad of\quad the\quad solute \end{matrix}$
Hence, $\begin{matrix} \triangle { T }_{ b } \\\ Elevation\quad in\quad boiling\quad point \end{matrix}\propto \begin{matrix} { x }_{ 2 } \\\ Mole\quad fraction\quad of\quad the\quad solute \end{matrix}$
Or, $\triangle { T }_{ b }=k{ x }_{ 2 }$
Where ‘k’ is proportionality constant.
Now, ${ x }_{ 2 }=\cfrac { { n }_{ 2 } }{ { n }_{ 2 }+{ n }_{ 1 } }$
Where ${ n }_{ 2 }$ is the number of moles of the solute and ${ n }_{ 1 }$ is the number of moles of the solvent.
For a dilute solution we can write the above equation as,
${ x }_{ 2 }=\cfrac { { n }_{ 2 } }{ { n }_{ 1 } }$ (We have ignored ${ n }_{ 2 }$ in comparison to ${ n }_{ 1 }$ in the denominator)
$\Rightarrow { x }_{ 2 }=\cfrac { { n }_{ 2 } }{ \dfrac { { w }_{ 1 } }{ { M }_{ 1 } } }$
Where ${ w }_{ 1 }$ is the mass of the solvent in the solution and ${ M }_{ 1 }$ is the molar mass of the solvent.
Therefore, $\triangle { T }_{ b }=k{ M }_{ 1 }\cfrac { { n }_{ 2 } }{ { w }_{ 1 } }$
If the mass of the solvent is equal to 1 Kg, then the fraction $\cfrac { { n }_{ 2 } }{ { w }_{ 1 } }$ will be equal to the molality of the solution.
For any solvent, the molar mass of that solvent is a fixed quantity, therefore we can write:
$k{ M }_{ 1 }={ K }_{ b }$
Where ${ K }_{ b }$ is called the molal elevation constant.
Therefore the final equation will be:
$\triangle { T }_{ b }={ K }_{ b }m$
Where m is the molality of the solution.
Hence the given question is explained.

Note: ${ K}_{ b }$ is also called ebullioscopic constant and can be calculated for a given solvent by using the formula:
${ K }_{ b }=\cfrac { R{ T }_{ 0 }^{ 2 } }{ 1000{ l }_{ v } } =\cfrac { { M }_{ 1 }R{ T }_{ 0 }^{ 2 } }{ 1000\triangle _{ vap }H }$
Where ${ T }_{ 0 }$ is the boiling point of the pure solvent, $\triangle _{ vap }H$ is the latent heat of evaporation per mole of the solvent, ${ M }_{ 1 }$ is the molar mass of the solvent, R is the gas constant and ${ l }_{ v }$ is the latent heat of evaporation per gram of the solvent.