Answer : The normal boiling point of ethanol will be,
or 
Explanation :
The Clausius- Clapeyron equation is :

where,
= vapor pressure of ethanol at
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:


Hence, the normal boiling point of ethanol will be,
or 
The intermolecular force that attracts two nonpolar molecules is London dispersion forces, which are also called induced dipole-induced
D. More collisions occur and the time required for the reaction decreases
This happens because according to collision theory, when energy (in this case, thermal energy) is applied to particles, they move/vibrate more quickly.