Question

The molar volume of liquid benzene (density = $0.877gm{L^{ - 1}}$) increases by a factor of 2750 as it vaporizes at $20^\circ C$ and that of liquid toluene (density = $0.867gm{L^{ - 1}}$) increases by a factor of 7720 at $20^\circ C$. A solution of benzene and toluene has a vapour pressure of 46 torr. Find the mole fraction of benzene at vapour pressure above the solution.

Answer 1

The mole fraction of benzene at vapour pressure above the solution has to be found. In order to do that we should be knowing the following formula and laws, such as Dalton's law, Raoult’s law, molecular formula, molar volume and ideal gas equation. By correctly substituting the given values in the equations we can obtain the mole fraction of benzene at vapour pressure above the solution.

Complete step by step answer:
From the question we can see that benzene has a density
= $0.877gm{L^{ - 1}}$
Now we have to calculate the molecular mass of benzene
$= 6 \times 12 + 6 \times 1 = 78$
- The above calculated molecular volume of benzene can be used to determine the molar volume of the liquid benzene, which is given by
${\text{molar volume = }}\dfrac{{{\text{molecular mass}}}}{{{\text{density}}}}$
Molar volume of benzene in liquid form
= $\dfrac{{78gmo{l^{ - 1}}}}{{0.877gm{l^{ - 1}}}}$
Molar volume of benzene in vapour phase
= $\dfrac{{78gmo{l^{ - 1}}}}{{0.877gm{l^{ - 1}}}} \times 2750 = 244583.8ml$
We can convert the above value by dividing by 1000.
Molar volume of benzene in vapour phase
= $\dfrac{{{\text{0}}{\text{.24458}}}}{{1000}} = 244.58L$

Now we have to calculate the molar volume of toluene.
Density of toluene is given as $0.867gm{L^{ - 1}}$
Now we have to calculate the molecular mass toluene
$= 6 \times 12 + 5 \times 1 + 1 \times 12 + 3 \times 1 = 92gmo{l^{ - 1}}$
Using this we can now calculate the molar volume of toluene
Molar volume of Toluene in liquid form
= $\dfrac{{92gmo{l^{ - 1}}}}{{0.867gm{l^{ - 1}}}}$
- Molar volume of Toluene in vapour phase
= $\dfrac{{92gmo{l^{ - 1}}}}{{0.867gm{l^{ - 1}}}} \times 7720 = 819192.6ml$

Now convert the above value by dividing by 1000.
Molar volume of Toluene in vapour phase
${\text{ = }}\dfrac{{{\text{0}}{\text{.81919}}}}{{1000}} = 819.19L$
To obtain the pressure of the benzene and toluene, we have to take the ideal gas equation
i.e. $PV = nRT$
For benzene,
$T = 298K$
$R = 0.0831$
$P = P_B^0 = \dfrac{n}{{{V_B}}}RT$
$P_B^0 = \dfrac{{1 \times 0.0831 \times 298}}{{244.58}}$
$P_B^0 = 0.098 atm$
$P_B^0 = 74.48 torr$
In order to convert the pressure from atm to torr, we have to multiply by 760
Because $1atm = 760 torr$
Similarly, for toluene
$P_T^0 = \dfrac{{1 \times 0.0831 \times 298}}{{819.19}}$
$P_T^0 = 22.04 torr$
Using Raoult’s law equation, we can calculate the pressure of benzene.
${P_B} = P_B^0{X_B} = 74.48{X_B}$
We can take, ${X_T} = 1 - {X_B}$ (using mole fraction)
${P_T} = P_T^0{X_T} = 22.04(1 - {X_B}) = 22.04 - 22.04{X_B}$
${P_M} = {P_B} + {P_T} = 74.48{X_B} + 22.04 - 22.04{X_B}$
Given, ${P_M} = 46 torr$
${X_B} = 0.456$
Using Dalton’s law, we can calculate the mole fraction of benzene in the vapour form.
${P_B} = {P_M}{X_B}$
${X_B} = \dfrac{{{P_B}}}{{{P_M}}}$
${X_B} = \dfrac{{74.48 \times 0.457}}{{46}}$
${X_B} = 0.73$.
Therefore, the mole fraction of benzene in vapour pressure above the solution is ${X_B} = 0.73$

Additional information: Some of the laws used, their definition and equation.
- Raoult’s law:
- According to Raoult’s law the partial vapor pressure of the solvent in a solution is equal to the product of vapour pressure of the pure solvent and its mole fraction in solution.
${P_{Solution}} = P_{Solvent}^\circ \times {X_{Solvent}}$
- Dalton’s law:
According to Dalton’s law, the total vapour pressure exerted is equal to the sum of the partial vapour pressure of individual gases.
${p_i} = {p_{total}}{x_i}$

Note: In order to convert atmospheric pressure into torr, we are multiplying the value of atmospheric pressure by 760 torr is the unit of pressure and one torr is equal to 133.33 Pa.
1 Torr = 1mm of Hg