$1{\text{mol}}$ benzene $\left( {{{\text{P}}^ \circ }_{{\text{benzene}}} = 42{\text{mm}}} \right)$ and $2{\text{mol}}$ toluene $\left( {{{\text{P}}^ \circ }_{{\text{toluene}}} = 36{\text{mm}}} \right)$ will have:
A. total pressure of $38{\text{mm}}$
B. mole fraction of vapor of benzene above liquid mixture is $\dfrac{7}{{19}}$
C. positive deviation from Raoult’s law.
D. negative deviation from Raoult’s law.
Answer
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Hint: The vapor pressure of the two different molecules is given in the question and the number of moles is also given. By using the Raoult’s formula, and substituting the given values in that formula. We can get the total pressure and also mole fraction of the two mixtures. Thus, we can find the correct answer.
Complete step by step answer:
The data given in the question are,
Vapor pressure of benzene, ${{\text{P}}^ \circ }_{{\text{benzene}}} = 42{\text{mm}}$
Vapor pressure of toluene, ${{\text{P}}^ \circ }_{{\text{toluene}}} = 36{\text{mm}}$
Number of moles of benzene$ = 1{\text{mol}}$
Number of moles of toluene$ = 2{\text{mol}}$
Raoult’s law: Raoult’s law states that the vapor pressure of a solution is directly proportional to the mole fraction of solvent. The solution which obeys Raoult’s law is called the ideal solution.
The solution of benzene and toluene is an ideal solution. Both of them have similar molecular structures. In this solution, the interactions between toluene and benzene are almost equal to benzene-benzene interactions and toluene-toluene interactions. This is because of the similarity in their molecular structures. Thus, we can say that there is no deviation in Raoult’s law. Therefore, options C and D are removed.
It is given that vapor pressure of benzene, ${{\text{P}}^ \circ }_{{\text{benzene}}} = 42{\text{mm}}$
Vapor pressure of toluene, ${{\text{P}}^ \circ }_{{\text{toluene}}} = 36{\text{mm}}$
Number of moles of benzene $ = 1{\text{mol}}$
Number of moles of toluene $ = 2{\text{mol}}$
Mole fraction of benzene $ = \dfrac{{{\text{number}}\;{\text{of}}\;{\text{moles}}\;{\text{of}}\;{\text{benzene}}}}{{{\text{number}}\;{\text{of}}\;{\text{total}}\;{\text{solution}}}} = \dfrac{{1{\text{mol}}}}{{1{\text{mol}} + 2{\text{mol}}}} = \dfrac{{1{\text{mol}}}}{{3{\text{mol}}}} = \dfrac{1}{3}$
Mole fraction of toluene $ = \dfrac{{{\text{number}}\;{\text{of}}\;{\text{moles}}\;{\text{of}}\;{\text{toluene}}}}{{{\text{number}}\;{\text{of}}\;{\text{total}}\;{\text{solution}}}} = \dfrac{{2{\text{mol}}}}{{1{\text{mol}} + 2{\text{mol}}}} = \dfrac{{2{\text{mol}}}}{{3{\text{mol}}}} = \dfrac{2}{3}$
Now we can calculate the total pressure. According to Raoult’s law,
${{\text{P}}_{{\text{total}}}} = {{\text{P}}^ \circ }_{{\text{benzene}}}{{\text{X}}_{{\text{benzene}}}} + {{\text{P}}^ \circ }_{{\text{toluene}}}{{\text{X}}_{{\text{toluene}}}}$
Substituting the values in equation, we get
$ \Rightarrow {{\text{P}}_{{\text{total}}}} = 42{\text{mm}} \times \dfrac{1}{3} + 36{\text{mm}} \times \dfrac{2}{3}$
On further simplifying,
$ \Rightarrow {{\text{P}}_{{\text{total}}}} = 14{\text{mm}} + 24{\text{mm = 38mm}}$
Thus, option A is correct.
Now considering option B, we have to calculate the mole fraction of benzene in terms of the ratio of vapor pressure of solvent solution.
According to Raoult’s law, ${{\text{P}}_{{\text{soln}}}} = {{\text{X}}_{{\text{solvent}}}}{{\text{P}}_{{\text{solvent}}}}$
Let ${{\text{X}}_{{\text{benzene}}}}$be the mole fraction of benzene. Here the solvent is benzene.
${{\text{X}}_{{\text{benzene}}}} = \dfrac{{{{\text{P}}_{{\text{benzene}}}}}}{{{{\text{P}}_{{\text{total}}}}}} = \dfrac{{{{\text{P}}^ \circ }_{{\text{benzene}}} \times {{\text{X}}^ \circ }_{{\text{benzene}}}}}{{{{\text{P}}_{{\text{total}}}}}} = \dfrac{{42{\text{mm}} \times \dfrac{1}{3}}}{{38{\text{mm}}}}$
On solving, we get
${{\text{X}}_{{\text{benzene}}}} = \dfrac{{14{\text{mm}}}}{{38{\text{mm}}}} = \dfrac{7}{{19}}$
$\therefore $ Both options A and B are correct.
Additional information-Vapor pressure of a liquid is much different in a solution than it is in pure liquid. The dissolved non-volatile solute lowers the vapor pressure of a solvent.
Note:
Vapor pressure is the pressure acted over a substance at which vapors are formed. When a plot of vapor pressure of solution, ${{\text{P}}_{{\text{soln}}}}$, against mole fraction of solvent, ${{\text{X}}_{{\text{solvent}}}}$, is represented, it gives a straight line with a slope of ${{\text{P}}_{{\text{solvent}}}}$.
Complete step by step answer:
The data given in the question are,
Vapor pressure of benzene, ${{\text{P}}^ \circ }_{{\text{benzene}}} = 42{\text{mm}}$
Vapor pressure of toluene, ${{\text{P}}^ \circ }_{{\text{toluene}}} = 36{\text{mm}}$
Number of moles of benzene$ = 1{\text{mol}}$
Number of moles of toluene$ = 2{\text{mol}}$
Raoult’s law: Raoult’s law states that the vapor pressure of a solution is directly proportional to the mole fraction of solvent. The solution which obeys Raoult’s law is called the ideal solution.
The solution of benzene and toluene is an ideal solution. Both of them have similar molecular structures. In this solution, the interactions between toluene and benzene are almost equal to benzene-benzene interactions and toluene-toluene interactions. This is because of the similarity in their molecular structures. Thus, we can say that there is no deviation in Raoult’s law. Therefore, options C and D are removed.
It is given that vapor pressure of benzene, ${{\text{P}}^ \circ }_{{\text{benzene}}} = 42{\text{mm}}$
Vapor pressure of toluene, ${{\text{P}}^ \circ }_{{\text{toluene}}} = 36{\text{mm}}$
Number of moles of benzene $ = 1{\text{mol}}$
Number of moles of toluene $ = 2{\text{mol}}$
Mole fraction of benzene $ = \dfrac{{{\text{number}}\;{\text{of}}\;{\text{moles}}\;{\text{of}}\;{\text{benzene}}}}{{{\text{number}}\;{\text{of}}\;{\text{total}}\;{\text{solution}}}} = \dfrac{{1{\text{mol}}}}{{1{\text{mol}} + 2{\text{mol}}}} = \dfrac{{1{\text{mol}}}}{{3{\text{mol}}}} = \dfrac{1}{3}$
Mole fraction of toluene $ = \dfrac{{{\text{number}}\;{\text{of}}\;{\text{moles}}\;{\text{of}}\;{\text{toluene}}}}{{{\text{number}}\;{\text{of}}\;{\text{total}}\;{\text{solution}}}} = \dfrac{{2{\text{mol}}}}{{1{\text{mol}} + 2{\text{mol}}}} = \dfrac{{2{\text{mol}}}}{{3{\text{mol}}}} = \dfrac{2}{3}$
Now we can calculate the total pressure. According to Raoult’s law,
${{\text{P}}_{{\text{total}}}} = {{\text{P}}^ \circ }_{{\text{benzene}}}{{\text{X}}_{{\text{benzene}}}} + {{\text{P}}^ \circ }_{{\text{toluene}}}{{\text{X}}_{{\text{toluene}}}}$
Substituting the values in equation, we get
$ \Rightarrow {{\text{P}}_{{\text{total}}}} = 42{\text{mm}} \times \dfrac{1}{3} + 36{\text{mm}} \times \dfrac{2}{3}$
On further simplifying,
$ \Rightarrow {{\text{P}}_{{\text{total}}}} = 14{\text{mm}} + 24{\text{mm = 38mm}}$
Thus, option A is correct.
Now considering option B, we have to calculate the mole fraction of benzene in terms of the ratio of vapor pressure of solvent solution.
According to Raoult’s law, ${{\text{P}}_{{\text{soln}}}} = {{\text{X}}_{{\text{solvent}}}}{{\text{P}}_{{\text{solvent}}}}$
Let ${{\text{X}}_{{\text{benzene}}}}$be the mole fraction of benzene. Here the solvent is benzene.
${{\text{X}}_{{\text{benzene}}}} = \dfrac{{{{\text{P}}_{{\text{benzene}}}}}}{{{{\text{P}}_{{\text{total}}}}}} = \dfrac{{{{\text{P}}^ \circ }_{{\text{benzene}}} \times {{\text{X}}^ \circ }_{{\text{benzene}}}}}{{{{\text{P}}_{{\text{total}}}}}} = \dfrac{{42{\text{mm}} \times \dfrac{1}{3}}}{{38{\text{mm}}}}$
On solving, we get
${{\text{X}}_{{\text{benzene}}}} = \dfrac{{14{\text{mm}}}}{{38{\text{mm}}}} = \dfrac{7}{{19}}$
$\therefore $ Both options A and B are correct.
Additional information-Vapor pressure of a liquid is much different in a solution than it is in pure liquid. The dissolved non-volatile solute lowers the vapor pressure of a solvent.
Note:
Vapor pressure is the pressure acted over a substance at which vapors are formed. When a plot of vapor pressure of solution, ${{\text{P}}_{{\text{soln}}}}$, against mole fraction of solvent, ${{\text{X}}_{{\text{solvent}}}}$, is represented, it gives a straight line with a slope of ${{\text{P}}_{{\text{solvent}}}}$.
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