Question 14 Marks
Vapour pressure of a liquid or a solution is the pressure exerted by the vapour in equilibrium with the liquid or solution at a particular temperature. It depends upon the nature of the liquid and temperature. The non-volatile solute in solution reduces the escaping tendency of the solvent molecules in the vapour phase because some of the solute particles occupy the positions of the solvent molecules on the liquid surface. The relative lowering of the vapour pressure of a solution containing a non$-$volatile solute is equal to the mole fraction of the solute in the solution. This is also known as Raoult's law. However, for solutions of volatile solutes, the vapour pressure of a component in a solution at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that pure component. The solutions in which each component obeys Raoult's law is called an ideal solution. For ideal solutions $\Delta H_{\text {mixing }}$ and $\Delta V_{\text {mixing }}$ are also zero. Practically no solution is ideal. A non$-$ideal solution is that solution in which solute and solvent molecules interact with one another with a different force than the forces of interaction between the molecules of the pure components. There are two types of non-ideal solutions, showing positive deviations and negative deviations from ideal behaviour. If for the two components$ A$ and $B$ , the forces of interaction between $A$ and $B$ molecules are less than the $A-A$ and $B-B$ interactions, the non-ideal solutions have positive deviations. On the other hand, if the forces of interaction between $A$ and $B$ molecules are more than the $A-A$ and $B-B$ interactions, the non-ideal solutions have negative deviations.
$i.$ What is the mole fraction of A in solution obeying result's low if the vapour pressure of a pure liquid A is $40 \ mm$ of $Hg$ at $300 \ K$ . The vapour pressure of this liquid in solution with liquid $B$ is $32$ mm of $Hg ?$
$ii.$ Vapour pressure of a solution of heptane $\&$ octane is given by the equation: $(1) P ( sol ).( mm Hg )=35+65 x$, where x is the mole fraction of heptane. Calculate the vapour pressure of pure octane.
$iii.$ What is the value of $\Delta V _{\text {mixing }}$ and $\Delta H _{\text {mixing }}$ for non$-$ideal solution showing negative deviation?
OR
$iii.$ Acetic acid $+$ pyridine, the mixture is an example of which type of solution?
$i.$ What is the mole fraction of A in solution obeying result's low if the vapour pressure of a pure liquid A is $40 \ mm$ of $Hg$ at $300 \ K$ . The vapour pressure of this liquid in solution with liquid $B$ is $32$ mm of $Hg ?$
$ii.$ Vapour pressure of a solution of heptane $\&$ octane is given by the equation: $(1) P ( sol ).( mm Hg )=35+65 x$, where x is the mole fraction of heptane. Calculate the vapour pressure of pure octane.
$iii.$ What is the value of $\Delta V _{\text {mixing }}$ and $\Delta H _{\text {mixing }}$ for non$-$ideal solution showing negative deviation?
OR
$iii.$ Acetic acid $+$ pyridine, the mixture is an example of which type of solution?
Answer
View full question & answer→$ (i) P _{ A }= x _{ A } \times P_A^{\circ}$
$32= x _{ A } \times 40$
$x _{ A }=\frac{32}{40}$
$x _{ A }=0.8$
$ii.$ For pure octane, $x=0$
$\therefore p(\text { sol })(mm Hg)=P(\text { octane })=35+65 \times 0=35 mm \text { of } Hg$
$iii.$ The value of $\Delta V _{\text {mixing }}$ and $\Delta H _{\text {mixing }}$ is negative.
$32= x _{ A } \times 40$
$x _{ A }=\frac{32}{40}$
$x _{ A }=0.8$
$ii.$ For pure octane, $x=0$
$\therefore p(\text { sol })(mm Hg)=P(\text { octane })=35+65 \times 0=35 mm \text { of } Hg$
$iii.$ The value of $\Delta V _{\text {mixing }}$ and $\Delta H _{\text {mixing }}$ is negative.
