Question

A solution prepared by dissolving 1.25 gm of oil of wintergreen (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31 ${}^{o}C$ . Determine the molar mass of this compound.
(Boiling point of pure benzene = 80.10 ${}^{o}C$ and ${{K}_{b}}$ for benzene = 2.53 ${}^{o}C$ Kg/mol)

Answer 1

Hint: By adding methyl salicylate the boiling point of benzene is going to increase from 80.10 ${}^{o}C$ to 80.31 ${}^{o}C$ . This is called elevation in boiling point.
There is a relationship between elevation in boiling point and weight of the samples and it is as follows.
\[Elevation\text{ }in\text{ }boiling\text{ }po\operatorname{int}(\Delta {{T}_{b}})=\dfrac{{{K}_{b}}\times 1000\times {{w}_{2}}}{M\times {{w}_{1}}}\]
$\Delta {{T}_{b}}$ = Elevation in boiling point.
${{w}_{1}}$ = weight of solvent
${{w}_{2}}$ = weight of solute
M = Mass of the solute.

Complete step by step answer:
- In the question it is given that by adding 1.25 g of oil of wintergreen the boiling point of benzene is changed from 80.10 ${}^{o}C$ to 80.31 ${}^{o}C$.
- We have to calculate the molar mass of the oil of wintergreen (solute).
- Substitute all the values in the below formula to get the molar mass of the solute.
\[Elevation\text{ }in\text{ }boiling\text{ }point\operatorname{ }(\Delta {{T}_{b}})=\dfrac{{{K}_{b}}\times 1000\times {{w}_{2}}}{M\times {{w}_{1}}}\]
$\Delta {{T}_{b}}$ = Elevation in boiling point = 80.31 - 80.10 = 0.21 ${}^{o}C$
${{w}_{1}}$ = weight of solvent = 99.0 gm
${{w}_{2}}$ = weight of solute = 1.25 gm
M = Mass of the solute = ?
${{K}_{b}}$ of benzene = 2.53 ${}^{o}C$ Kg/mol
- Therefore
\[\begin{align}
  & Elevation\text{ }in\text{ }boiling\text{ }po\operatorname{int}(\Delta {{T}_{b}})=\dfrac{{{K}_{b}}\times 1000\times {{w}_{2}}}{M\times {{w}_{1}}} \\
 & \Rightarrow M=\dfrac{{{K}_{b}}\times 1000\times {{w}_{2}}}{\Delta {{T}_{b}}\times {{w}_{1}}} \\
 & \Rightarrow M=\dfrac{(2.53\times {{10}^{3}})(1000)(1.25)}{(0.21)(99)} \\
 & \Rightarrow M=152.11\times {{10}^{3}}g/mol \\
\end{align}\]

Thus the molar mass of the compound is $152.11\times {{10}^{3}}g/mol$

Note: If the boiling point of a solvent is increased by the addition of a solute then it is called elevation in boiling point. The reason for the elevation of boiling point is because the added solute forms a layer on the surface of the solvent and stops the evaporation rate of the solvent.