Question

The degree of dissociation of $Ca{\left( {N{O_3}} \right)_2}$ in a dilute aqueous solution, containing $7{\text{ }}g$ of the salt per $100g$ of water at $100^\circ C\;$ is $70\%$. If the vapour pressure of water at $100^\circ C\;$ is $760mm$ , calculate the vapour pressure of the solution.

Answer 1

When there is a liquid placed in a vessel which is getting heated continually, then the liquid molecules start moving in different directions at different speed. This occurs as the kinetic energies of each particle is different in that liquid because of the heating. The energy of the molecules rises, and these molecules become lighter and thus occupy the liquid surface. This process is known as ‘evaporation’. The molecules which can be seen on the liquid surface are called ‘vapor

Complete step-by-step solution: When we define vapour pressure, we can identify it as the pressure exerted by a vapour, at a specific temperature, with its condensed phases (solid or liquid) in a system which is close. The propensity of particles to escape from the liquid (or a solid) is known to be related. A material which has high vapour pressure is known as a volatile material. It is to be noted that the pressure which is exhibited above a liquid surface by the vapour is known as the vapour pressure of that liquid.
Here in the above situation,
We need to find the moles of $Ca{\left( {N{O_3}} \right)_2}$and water
To find vapour pressure = ${P_o} - P/{P_o}{\text{ }} = {\text{ }}n/n{\text{ }} + {\text{ }}N$
We will take one mole of $Ca{\left( {N{O_3}} \right)_2}$
Degree of dissociation of $Ca{\left( {N{O_3}} \right)_2}$ $= {\text{ }}70/100{\text{ }} = {\text{ }}0.7$
Ionization of $Ca{\left( {N{O_3}} \right)_2}$ can be represented as $= {\text{ }}Ca{\left( {N{O_3}} \right)_2} \rightleftharpoons C{a^{2 + }}{\text{ + }}2N{O^{ - 3}}$
At start $100$
At equilibrium $1{\text{ }} - 0.7{\text{ }}0.7{\text{ }}2{\text{ }} \times 0.7$
Therefore, total number of moles in the solution at equilibrium
$= {\text{ }}\left( {1{\text{ }} - 0.7} \right){\text{ }} + {\text{ }}0.7{\text{ }} + {\text{ }}2{\text{ }} \times 0.7{\text{ }} = {\text{ }}2.4$
Number of moles when the solution contains one gm of calcium nitrate instead of one mole of the salt
$= {\text{ }}2.4/164{\text{ }}\left( {164{\text{ }}is{\text{ }}the{\text{ }}mol.{\text{ }}wt.{\text{ }}of{\text{ }}Cal.{\text{ }}nitrate} \right)$Thus, number of moles of the solute in the solution containing $7{\text{ }}g$of salt
$n{\text{ }} = {\text{ }}2.4/164 \times 7{\text{ }} = {\text{ }}0.102$
$No.{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}water{\text{ }}\left( N \right){\text{ }} = {\text{ }}Weight{\text{ }}of{\text{ }}water/Mol.{\text{ }}wt.{\text{ }}of{\text{ }}water{\text{ }} = {\text{ }}100/18{\text{ }} = {\text{ }}5.55$Applying Raoult’s law
${P_o} - P/{P_o}{\text{ }} = {\text{ }}n/n{\text{ }} + {\text{ }}N$
$\Rightarrow 760{\text{ }}-p/760{\text{ }} = {\text{ }}0.102/0.102{\text{ }} + {\text{ }}5.55$
$\Rightarrow 760{\text{ }}-{\text{ }}p/760{\text{ }} = {\text{ }}0.0180$
$\Rightarrow p = {\text{ }}p{\text{ }} = {\text{ }}760{\text{ }}-{\text{ }}\left( {760{\text{ }} \times {\text{ }}0.0180} \right)$
$\Rightarrow p = {\text{ }}746.3{\text{ }}mm{\text{ }}Hg$

Note: Equilibrium is a stage reached when the rate of evaporation is equal to the rate of condensation. The rate of evaporation is dependent on various factors such as the nature of the liquid, thus when a liquid is heated the particles move faster and thus we can see an increase in the vapour pressure. The vapour pressure of the liquid is known to increase when there is an increase in its temperature seen . The molecules of the liquid have higher energy at higher temperatures.