Question

The relative lowering of vapour pressure of 0.2 molal solution in which the solvent in benzene is
(A) $15.6 \times {10^{ - 4}}$
(B) $15.6 \times {10^{ - 3}}$
(C) $15.6 \times {10^{ - 1}}$
(D) $0.05$

Answer 1

Hint: We know that there are two types of solutes: volatile and non-volatile. When we boil a mixture of both solutes, the non-volatile remains in the given phase, causing the vapour to produce at a lower pressure than when there is only fixed vapour. This decrease in pressure is referred to as a relative decrease in vapour pressure.

Complete Step by Step Solution:
Here in the question given that,
m (molality) = 0.2
We have to find the relative lowering in vapour pressure in the solvent of benzene,

As taking according to Raoult’s law,
lowering of vapour pressure = Mole fraction of Solute
From which we say that,
$m(molality) = \dfrac{{{X_{solute}} \times 1000}}{{(1 - {X_{solute}}){{(GMW)}_{solvent}}}}$
Here, GMW stands for Gram Molecular Weight,
Gram Molecular Weight of benzene = $(6 \times 12) + (6 \times 1) = 78$
By putting all the values in above equation,
$0.2 = \dfrac{{{X_{solute}} \times 1000}}{{(1 - {X_{solute}})78}}$
By using cross multiplication here, we get,
$15.6 - 15.6{X_{solute}} = 1000{X_{solute}}$
By doing further solution we get,
$15.6 = 1015.6{X_{solute}}$
As taking the value of this by doing calculation we get,
${X_{solute}} = 0.0156 = 15.6 \times {10^{ - 3}}$
As we mentioned above that,
The Average lowering of vapour pressure = Mole fraction of Solute = ${X_{solute}} = 0.0156 = 15.6 \times {10^{ - 3}}$
As from this we can say that, Relative lowering of vapour pressure $ = 15.6 \times {10^{ - 3}}$
Hence, the correct option is (B).

Note: As relative lowering in vapour pressure is the topic of colligative properties in which most of the numerical questions are present in it of the whole chapter and out of all that topics relative lowering in vapour pressure have the most number of numerical questions putted in from the colligative properties.