Question

The normal boiling point of toluene is nearly ${110.7^ \circ }\;C$ and its boiling point elevation constant is $3.32\;K\;kg\;mo{l^{ - 1}}$. The enthalpy of vaporization of toluene is nearly
(A) $17\;KJ\;mo{l^{ - 1}}$
(B) $21\;KJ\;mo{l^{ - 1}}$
(C) $51\;KJ\;mo{l^{ - 1}}$
(D) $68\;KJ\;mo{l^{ - 1}}$

Answer 1

Hint: The enthalpy of vaporisation and the boiling point elevation constant are related by the equation derived from the references from Raoult’s law. We need to find the enthalpy of evaporation through that relationship. All other parameters are given such as boiling point and the boiling point elevation constant.

Complete step by step solution: We know that the relation between molal elevation constant ${K_b}$ and enthalpy of vaporisation is given by
${K_b} = \dfrac{{MRT_b^2}}{{1000{E_{vap}}}}$ where ${K_b}$ is the molal elevation constant
M is the molar mass of the molecule
${T_b}$ is the boiling point
R is the universal gas constant
${E_{vap}}$ is the enthalpy of evaporation
So we have been provided with the following information
${K_b}$=$3.32\;K\;kg\;mo{l^{ - 1}}$
${T_b} = {110.7^ \circ }\;C$$ = 110.7 + 273 = 383.7\;K$
$R = 8.314\;Nm\;{K^{ - 1}}$
Boiling point elevation constant is given by
 ${K_b} = \dfrac{{MRT_b^2}}{{1000{E_{vap}}}}$
So putting the given values in the equation we get
 ${E_{vap}} = \dfrac{{92 \times 8.314 \times {{(383.7)}^2}}}{{1000 \times 3.32}}$
 $ \Rightarrow {E_{vap}} = 33.8 \approx 34\;KJ\;mo{l^{ - 1}}$
Therefore the enthalpy of vaporisation is approximately $34\;KJ\;mo{l^{ - 1}}$

Hence, none of the options is correct.

Note: Here the temperatures are present in degree celsius which should be converted to kelvin before applying into the formula. Also, the enthalpy of evaporation of a liquid is proportional to the thermodynamic temperature at which the liquid boils. This is termed as Trouton’s rule. In the given concept it is useful to write the concentration in terms of molality instead of mole fraction while the derivation of the equation.