Question

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

Answer 1

Answer: 5

$0.1$ mole ionic salt in $1.8kg$ water at $35^{\circ }C$
Vapour pressure of solution $=59.724mm$ of $Hg$
Vapour pressure of pure $H_{2}O=60.000mm$ 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
$\frac{P^{\circ }-P_s}{P_s}=\frac{\text{ moles of non }-\text{ volatileparticles }}{\text{ moles of water }}$
$\frac{60.000-59.724}{59.724}=\frac{0.01+0.09x}{100}$
$(0.276)\times 100=0.59274+(0.59274\times 9)x$
$27.6-0.59274=(0.59274\times 9)x$
$\Rightarrow x\simeq \frac{27}{0.6\times 9}=5$