Question

The lowering of vapour pressure of ${\text{0}}{\text{.1M}}$ aqueous solutions of ${\text{NaCl}}$ , ${\text{CuS}}{{\text{O}}_{\text{4}}}$ and ${{\text{K}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ are:
A.All equal
B.In the ratio of $1:1:1.5$
C.In the ratio of $3:2:1$
D.In the ratio of $1.5:1:2.5$

Answer 1

Relative lowering of vapour pressure is a colligative property as it is independent of the nature of the solute but is dependent on the moles of that solute.
For substances undergoing association or dissociation, the relative lowering of vapour pressure is equal to the van’t Hoff factor multiplied by mole fraction of solute.

Complete step by step answer:
The relation between the relative lowering of vapour pressure of a solution and the mole fraction of the non-volatile solute in the solution for substances undergoing association or dissociation is given by Raoult’s law and the expression for this law is given by:
$\dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}^{\text{0}}}}}{\text{ = i}}{{\text{x}}_2}$
where ${{\text{p}}^{\text{0}}}$ denotes the vapour pressure of the pure solvent, ${{\text{p}}_{\text{s}}}$ denotes the vapour pressure of the solution, ‘i’ denotes the van’t Hoff factor.
So, the term ${{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}$ represent the lowering of vapour pressure and $\dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}^{\text{0}}}}}$ represent the relative lowering of vapour pressure and ${{\text{x}}_{\text{2}}}$ represent the mole fraction of that solute.
Now, ${\text{0}}{\text{.1M}}$ aqueous solutions means ${\text{0}}{\text{.1}}$ moles in one litre of solution and so mole fraction ${{\text{x}}_{\text{2}}} = 0.1$
Now, van’t Hoff factor is equal to the number of solute particles.
So, for ${\text{NaCl}}$ , the van’t Hoff factor ‘i’ is equal to 2 as it dissociates into one sodium ion and one chloride ion.
${\text{NaCl}} \to {\text{N}}{{\text{a}}^{\text{ + }}}{\text{ + C}}{{\text{l}}^{\text{ - }}}$
For ${\text{CuS}}{{\text{O}}_{\text{4}}}$ , van’t Hoff factor ‘i’ is equal to 2 as it dissociates into one copper (II) ion and one sulphate ion.
${\text{CuS}}{{\text{O}}_{\text{4}}} \to {\text{C}}{{\text{u}}^{{\text{2 + }}}}{\text{ + S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$
For ${{\text{K}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ , van’t Hoff factor ‘i’ is equal to 3 as it dissociates into two potassium ions and one sulphate ion.
${{\text{K}}_2}{\text{S}}{{\text{O}}_{\text{4}}} \to 2{{\text{K}}^{\text{ + }}}{\text{ + S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$
So, the lowering of vapor pressure for ${\text{NaCl}}$ will be
$\dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}^{\text{0}}}}}{\text{ = 2}} \times {\text{0}}{\text{.1 = 0}}{\text{.2}}$
For ${\text{CuS}}{{\text{O}}_{\text{4}}}$, it will be \dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}^{\text{0}}}}}{\text{ = 2}} \times {\text{0}}{\text{.1 = 0}}{\text{.2}}\ For ${{\text{K}}_2}{\text{S}}{{\text{O}}_{\text{4}}}$ , it will be \[\dfrac{{{{\text{p}}^{\text{0}}}{\text{ - }}{{\text{p}}_{\text{s}}}}}{{{{\text{p}}^{\text{0}}}}}{\text{ = 3}} \times {\text{0}}{\text{.1 = 0}}{\text{.3}}
So, the required ratio is $0.2:0.2:0.3$ or $1:1:1.5$

So the correct option is (2).

Note:
If there is dissociation of the solute in the solution, then ‘i’ is greater than 1 and if there is association of the solute in the solution, then ‘i’ is less than 1.
The van’t Hoff factor ‘i’ is equal to the observed colligative property divided by the calculated colligative property.