The standard molar enthalpy of vaporisation $\left(\Delta_{\text {vap }} H^{\circ}\right)$ of $A, B$ and $C$…
The standard molar enthalpy of vaporisation $\left(\Delta_{\text {vap }} H^{\circ}\right)$ of $A, B$ and $C$ liquids is $23.3,41$ and $29 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The correct order of diople-dipole attractive forces in these liquids is
$B>C>A$
$B>A>C$
$A>C>B$
$A>B>C$
Solution
( The enthalpy of vaporisation is the amount of heat that must be added to a liquid substance to transform it into a gas. Dipole-Dipole interactions are intermolecular force of attraction between polar molecules. Stronger the dipole-dipole interactions, higher will be the heat energy required to break these interactions and vapourise the liquid. Since, the liquids $A, B$ and $C$ have molar enthalpy of vaporisation as $23.3,41$ and $29 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the order of dipole-dipole attractive interactions will be $B>C>A$.