Question

At $293K$ , $ethyl$ $acetate$ has vapour pressure of $72.8torr$ of $Hg$ and $ethyl{\text{ }}propionate\;$ has pressure of $27.7torr$ of $Hg$ . Assuming their mixture to obey Raoult’s law, determine the vapour pressure of a mixture containing $25g$ of $ethyl\;acetate$ and $50g$ of $ethyl{\text{ }}propionate\;$ .
(A) $44.4$
(B) $38.56$
(C) $57.28$
(D) $62.84$

Answer 1

Hint : To find the solution of the above question first find the number of moles of $ethyl$ $acetate$ and $ethyl{\text{ }}propionate\;$ and then the mole fraction of these two compounds. Then put these values in Raoult's law for vapour pressure.

Complete Step By Step Answer:
Raoult’s law states that a solvent’s partial vapour pressure in a solution is equal or the same as the vapour pressure of the pure solvent multiplied by its mole fraction in the solution. This means that the freezing and boiling points of an ideal solution are respectively depressed and elevated relative to that of the pure solvent by an amount proportional to the mole fraction of the solute.
Mathematically it is represented by –
${P_{solution}} = {X_{solvent}}.P_{solvent}^{^ \circ }$
Where,
${P_{solution}}$ is the vapour pressure of the solution
${X_{solvent}}$ is the mole fraction of the solvent
$P_{solvent}^{^ \circ }$ is the vapour pressure of the pure solvent.
The above equation is for a single liquid solution.
For two liquids it is given by
$P = {x_{A.}}P_A^ \circ + {x_B}P_B^ \circ$
Here A and B stand for two different liquids.
As mentioned in the hint that we must find the number of each compounds given to us
Number of moles is given by the mass ratio to the molar mass of the compound.
Moles of $ethyl$ $acetate$ =
${n_A} = \dfrac{{25}}{{88}} = 0.284$
Moles of $ethyl{\text{ }}propionate\;$ =
${n_B} = \dfrac{{50}}{{102}} = 0.49$
Now we will find out the mole fraction of each compound. It is the ratio of units of the amount of a constituent divided by the total amount of all constituents.
Mole fraction of $ethyl$ $acetate$ =
${x_A} = \dfrac{{0.284}}{{0.284 + 0.490}} = 0.367$
Mole fraction of $ethyl{\text{ }}propionate\;$ =
${x_B} = \dfrac{{0.490}}{{0.284 + 0.490}} = 0.633$
Now putting these value in the formula
$P = {x_{A.}}P_A^ \circ + {x_B}P_B^ \circ$
$P = 0.367 \times 72.8 + 0.633 \times 27.7$
$= 44.24torr$
Therefore our answer is option A.

Note :
Vapour pressure is exhibited by all solids and liquids and it depends on the type of liquid and the temperature. It is the measure of tendency of a material to change into the gaseous or vapour state, and increase with the temperature.