Question

The boiling point of liquid A, B, C and D are ${80^ \circ }C$, ${60^ \circ }C$, ${90^ \circ }C$ and ${100^ \circ }C$ respectively. Which will show the highest vapour pressure at room temperature?
(A) A
(B) B
(C) C
(D) D

Answer 1

To solve this question, we should know the definition of boiling point and vapour pressure. Knowing the effect of a boiling point on vapour pressure would help us to come to a conclusion within no time.

Complete step by step solution:
The boiling point can be defined as the point at which the pressure exerted by the surroundings on the liquid is equal to the pressure exerted by the vapour of the liquid.
Vapour pressure is the pressure exerted by the vapour of the liquid.
Boiling point is inversely proportional to vapour pressure. So, higher the boiling point lower will be the vapour pressure.
Option A is ${80^ \circ }C$, option B is ${60^ \circ }C$, option C is ${90^ \circ }C$ and Option D is ${100^ \circ }C$.
Since, option D is ${100^ \circ }C$, which has a higher boiling point among the other options. Thus, it is less vapour pressure than other given options.
Option C is ${90^ \circ }C$. The boiling point is a bit lower than that of liquid D. The vapour pressure of liquid C is a bit higher vapour pressure than the vapour pressure of liquid D.
Option A is ${80^ \circ }C$. Liquid B has a lower boiling point than liquid D and C. Therefore liquid A has higher vapour pressure than liquid D and C.
Option B is ${60^ \circ }C$. Liquid B has the least boiling in comparison to liquid A, C, D. From the above discussion, we know that, with an increase in boiling point, there will be a decrease in vapour pressure.

Thus, liquid B has a higher vapour pressure than liquid A, C and D.

Note: Evaporation rate of a liquid depends on vapour pressure. The ease of vapour formation is based on boiling point. The boiling point will be higher when the intermolecular bond between atoms is very strong. So if the boiling point is strong the ease of vapour formation becomes difficult and vice versa.