Question

An aqueous solution is prepared by dissolving $0.1 \,mol$ of an ionic salt in $1.8\, kg$ of water at $35^{\circ} C$. The salt remains $90 \%$ dissociated in the solution. The vapour pressure of the solution is $59.724\, mm$ of Hg. Vapor pressure of water at $35^{\circ} C$ is $60.000\, mm$ of Hg. The number of ions present per formula unit of the ionic salt is_____

Answer 1

$0.1$ mole ionic salt in $1.8 \,kg$ water at $35^{\circ} C$
Vapour pressure of solution $=59.724\, mm$ of $Hg$
Vapour pressure of pure $H _2 O =60.000\, mm$ of $Hg$
Let the number of ions present per formula unit of the ionic salt be ' $x$
'

Moles of water $=\frac{1.8 \times 10^3}{18}=100$ moles
Relative lowering of vapour pressure $\frac{ P ^{\circ}- P _s}{ P ^{\circ}}=$ Mole fraction of non - volatile particles
$ \frac{ P ^{\circ}- P _s}{ P _{ s }}=\frac{\text { moles of non }-\text { volatileparticles }}{\text { moles of water }}$
$ \frac{60.000-59.724}{59.724}=\frac{0.01+0.09 x }{100} $
$ (0.276) \times 100=0.59274+(0.59274 \times 9) x$
$ 27.6-0.59274=(0.59274 \times 9) x$
$ \Rightarrow x \simeq \frac{27}{0.6 \times 9}=5$