Question

The vapour pressure of methanol at a certain temperature is $1atm$ by adding a small amount of ethyl acetate the vapour pressure of the solution is found to be $684mm$. The relative lowering of vapour pressure is:
A.$0.1$
B.$0.9$
C.$76$
D.$0.694$

Answer 1

Hint: On adding a non-volatile solute to a pure solvent, the vapour pressure of the resulting solution is found to be less than that of the pure solvent, this is known as relative lowering of vapour pressure. Mathematically, it is the ratio of the lowering in vapour pressure (that is, the difference in vapour pressure of the pure solvent and the solution) and the vapour pressure of the pure solvent.

Complete step by step answer:
The relative lowering in vapour pressure can be calculated by using the formula,
$\dfrac{{\Delta p}}{{{p^0}}} = \dfrac{{{p^0} - {p_s}}}{{{p^0}}}$
The vapour pressure of the pure solvent is denoted by ${p^0}$. The pure solvent is methanol which has a vapour pressure of $1atm$, so, ${p^0} = 1atm$. We will convert the unit of pressure from atm to mmHg.
As, $1atm = 760mm$
So, ${p^0} = 760mm$
${p_s}$ is the vapour pressure of the solution, so, ${p_s} = 684mm$.
So, $\dfrac{{\Delta p}}{{{p^0}}} = \dfrac{{{p^0} - {p_s}}}{{{p^0}}}$
$ \Rightarrow \dfrac{{\Delta p}}{{{p^0}}} = \dfrac{{760 - 684}}{{760}}$
$ \Rightarrow \dfrac{{\Delta p}}{{{p^0}}} = \dfrac{{76}}{{760}}$
$ \Rightarrow \dfrac{{\Delta p}}{{{p^0}}} = 0.1$
Therefore, the relative lowering in vapour pressure is $0.1$

Hence option A is correct.

Note:
In case of pure solvent the surface of the liquid consists of only the molecules of pure solvent and an equilibrium exists between the molecules escaping into the vapour phase and those existing as the solvent. So, the molecules in the vapour phase exert a pressure on the surface of the solvent, known as the vapour pressure of the pure solvent. On adding a non-volatile solute, the surface of the solution is occupied by both the solute and the solvent molecules, which reduces the number of solvent molecules escaping into the vapour phase, which in turn reduces the pressure exerted by the vapour phase. Hence, the relative lowering in vapour pressure takes place.