Question

The boiling point elevation constant for toluene is \[3.32K\,kg\,mo{l^{ - 1}}\] and the normal boiling point of toluene is \[{110^0}C\] . The enthalpy of evaporation of toluene would by nearly
A. \[17.0\,kJ\,mo{l^{ - 1}}\]
B. \[34.0\,kJ\,mo{l^{ - 1}}\]
C. \[51.0\,kJ\,mo{l^{ - 1}}\]
D. \[68.0\,kJ\,mo{l^{ - 1}}\]

Answer 1

Hint: The enthalpy of vaporisation is the amount of energy which is required by the liquid which helps it to transform into the gaseous state of matter. The enthalpy which helps the solid to directly turn into a vapour state is known as sublimation.

Formula used: Since the enthalpy of vaporisation is to be calculated and we have the boiling point of toluene, the formula to be used is:
\[{K_b} = \dfrac{{RT_b^2M}}{{\Delta {H_{vap}}}}\] where,
\[{K_b}\] denotes the ebullioscopic constant which represents the boiling measurement and its SI unit is \[K\,kg\,mo{l^{ - 1}}\] .
\[R\] represents the ideal gas constant which has a value of \[8.314 \times {10^{ - 3}}\] .
\[M\] represents the molar mass of the solvent, in this question the molar mass of toluene is 92.
\[\vartriangle {H_{vap}}\] represents the enthalpy of vaporisation and its value is measured in J/Kg.
\[{T_b}\] represents the boiling point of the solvent. The value \[{T_b}\] is measured in Kelvin.

Complete Step by Step Solution:
Enthalpy of vaporisation is defined as the amount of energy that is required by the liquid to transform itself from liquid to gaseous form. When a solid skips its liquid form and directly turns into a vapour form, then that process is termed sublimation.
The enthalpy of vaporisation can be calculated by the given formula:

\[{K_b} = \dfrac{{RT_b^2M}}{{\Delta {H_{vap}}}}\] , so after the placement of values, we will obtain
\[3.32 = \dfrac{{8.314\,x\,{{10}^{ - 3}}x\,T_b^2\,x\,92}}{{\Delta {H_{vap}}}}\]
Since the temperature is to be taken in kelvin:
\[{110^o}C + 273.15K = 383.15K\] , therefore
\[3.32 = \dfrac{{8.314\,x\,{{(383.15)}^2}\,x\,92}}{{\Delta {H_{vap}}\,x\,1000}}\]

 \[\Delta {H_{vap}} = \dfrac{{112.6}}{{3.32}}\,kg\,mo{l^{ - 1}}\]
 \[ = 33.9KJ\]

Hence option B is the correct answer

Note: There is a very close correlation between the temperature at which the liquid boils and its enthalpy of evaporation. They both are directly proportional to each other and it's also known as Trouton’s rule.