Question

Benzene and toluene form two ideal solutions A and B at 313 K, Solution A (total pressure PA) contains equal mole of toluene and benzene. Solution B contains equal masses of both (total pressure PB). The vapour pressure of benzene and toluene are 160 and 60 mm of Hg respectively at 313 K. Calculate the value of \[{P_A} + {\text{ }}{P_B}\] (Round-off the answer).

Answer 1

Hint: According to Raoult's law, the vapour pressure of a component at a given temperature equals the mole fraction of that component times the vapour pressure of that component in a pure state.

Complete step by step answer:
For the solution A, the mole fractions for toluene and benzene are the same i.e. 0.5 in each case.
The partial pressure of toluene in a vapour phase is:
${P_T} = {P_T}^0 = 60mmHg \times 0.5 = 30mmHg$
Here, \[{P_T}^0\] refers to the vapour pressure of pure toluene while \[{X_T}\] refers to the mole fraction of toluene in liquid phase.
Now, the partial pressure of benzene in a vapour phase is:
${P_B} = {P_B}^0{X_B} = 160mmHg \times 0.5 = 80mmHg$
Here, \[{P_B}^0\] refers to the vapour pressure of pure benzene while \[{X_B}\;\] refers to the mole fraction of benzene in liquid phase.
Therefore, total pressure in case of solution A becomes:$$$$
${P_A} = {P_T} + {P_B} = 30mmHg + 80mmHg = 110mmHg$
For solution B, masses of the toluene and benzene are the same.
The molar mass of benzene is 78.11 g/mol and molar mass of toluene is 92.14 g/mol.
Let us suppose that 78.11 g of benzene and 78.11 g of toluene are present in the solution mixture.
This is equivalent to 1 mole of benzene and 0.847 mole of toluene.
The mole fraction of benzene is 0.5412 and mole fraction of toluene is 0.4588.
The partial pressure of toluene in the vapour phase is:
${P_T} = {P_T}^0{X_T} = 60mmHg \times 0.4588 = 27.53mmHg$
Here, \[{P_T}^0\]​refers to the vapour pressure of pure toluene while \[{X_T}\]refers to the mole fraction of toluene in the liquid phase.
The partial pressure of benzene in vapour phase is:${P_B} = {P_B}^0{X_B} = 160mmHg \times 0.5412 = 86.592mmHg$
Here, \[{P_B}^0\]refers to the vapour pressure of pure benzene while \[{X_B}\;\]​refers to the mole fraction of benzene in the liquid phase.
Therefore, the total pressure becomes: ${P_B} = {P_T} + {P_B} = 27.53mmHg + 86.592mmHg = 114.122mmHg$
Thus, the value of ratio i.e. $\dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{114.122mmHg}}{{110mmHg}} = 0.964$

Note: Dalton's law states that the total pressure is the sum of individual pressures at a constant temperature. This empirical law was invented by John Dalton in 1801. This law is basically correlated to the ideal gas law.