Question

The vapour pressures of ethanol and methanol are $44.5mm$ and 88.7mm$$$$Hg respectively. An ideal solution is formed at the same temperature by mixing $60{\text{ }}g$ of ethanol with $40{\text{ }}g\,$ of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Answer 1

The vapour pressure is the sum of the partial pressure of individual two components. The partial pressure of each component is equal to the product of mole fraction and partial pressure of pure component.

Complete step by step answer:
To find out the total vapor pressure first it is important to know information about the mole fraction, number of moles and partial pressure.
Mole fraction is defined as the number of the constituents that are divided by the total number of the moles present in the given mixture. The number of moles can be found by dividing the mass or weight of the compound to the molar mass of that compound.
Now, you can find the partial pressure by the help of Raoult’s law which states that partial pressure of each component is equal to the vapor pressure of the pure component multiplied by the mole fraction.
Given,
$P{^\circ _{ethanol}} = 44.5mmHg,{\text{ P}}{^\circ _{methanol}} = 88.7mmHg \\\ {W_{ethanol}} = 60g,{\text{ }}{{\text{W}}_{methanol}} = 40g \\\$

$\therefore$Molar mass of ethanol $\left( {{C_2}{H_S}OH} \right) = 46g$
Number of moles of ethanol $= \dfrac{{weight\,of\,ethanol}}{{molar\,mass\,of\,ethanol}}$
$= \dfrac{{60}}{{46}}$
$= 1.304$
Molar mass of methanol $\left( {C{H_3}OH} \right) = 32g$
Number of moles of methanol $\left( {C{H_3}OH} \right) = \dfrac{{Weight\,of\,methanol}}{{Molar\,mass\,of\,methanol}}$
$= \dfrac{{40}}{{32}}$
$= 1.25$
To find the mole fraction of ethanol and methanol, we have
Mole fraction $\left( X \right) = \dfrac{{number{\text{ of moles of one component}}}}{{total{\text{ number of moles of solution}}}}$
By using the formula, we get
$'X{'_A}$Mole fraction of ethanol $= \dfrac{{{n_{ethanol}}}}{{{n_{ethanol}} + {n_{methanol}}}}$
$= \dfrac{{1.304}}{{1.304 + 1.25}}$
$\therefore 'X{'_A}$mole fraction of ethanol$= 0.5107$
Similarly, $'X{'_B}$mole fraction of methanol $= \dfrac{{{n_{methanol}}}}{{{n_{methanol}} + {n_{ethanol}}}}$
$= \dfrac{{1.25}}{{1.25 + 1.304}}$
$\therefore 'X{'_B}$mole fraction of methanol$= 0.4893$
To find the partial pressure of ethanol and methanol, we have
Partial pressure of ethanol $\left( {{P_A}} \right) = P{^\circ _A}{X_A}$
$= 0.5107 \times 44.5$
$= 22.73mm$
And Partial pressure of methanol $\left( {{P_B}} \right) = P{^\circ _B}{X_B}$
$= 0.4893 \times 88.7$
$= 43.40mm$
Total vapour pressure of solution $= {P_A} + {P_B}$
$= 22.73 + 43.40$
$= 66.13mmHg$
To find the mole fraction of methanol in vapour phase, we have
Mole fraction of methanol in vapour$= \dfrac{{{\text{Partial pressure of methanol}}}}{{Total{\text{ pressure}}}}$
By putting the value of partial pressure of methanol and total pressure, we get
$\begin{gathered} = \dfrac{{43.40}}{{66.13}} \\\ = 0.6563 \\\ \end{gathered}$

Note:
The solutions which follow Raoult’s law over the entire range of concentration are called ideal solutions. For such solutions, observed vapour pressure is the same as the vapour pressure calculated using expression for Raoult’s law. Therefore, change in enthalpy of solution is equal to zero and change in volume of solution is equal to zero. Thermodynamic data for such solutions can be summarised by knowing them at molecular level.