Question

A mixture of volatile components. A and B has total vapour pressure in\[\left( {{\mathbf{Torr}}} \right)\], as \[{\mathbf{P}} = {\mathbf{254}} - {\mathbf{119}}{{\mathbf{x}}_A}\], Where \[{{\mathbf{x}}_A}\]is the mole fraction of A in the mixture. Hence \[{{\mathbf{P}}^o}_A\] and \[{{\mathbf{P}}^o}_B\]respectively are in \[\left( {{\mathbf{Torr}}} \right)\] :
A.254,119
B.119,254
C.135,254
D.154,119

Answer 1

Hint:The Raoult’s law gives the total vapour pressure of a given mixture of two liquids or any amount of liquids present in the solution. The equation of Raoult’s law can give us an accurate idea to answer this question.

Complete step by step answer:
Raoult’s law states that the partial vapour pressure of each solute is directly proportional to the total vapour pressure of the solution and the constant for this proportionality equation is the mole fraction of that solute.
${P_A}\alpha {P^o}_A$ ;
Where, ${P_A}$ is partial vapour pressure of component A on the solution;
${P^o}_A$ is pure vapour pressure of the component A.
${P_A} = {x_A}{P^o}_A - (1)$;
Now for partial vapour pressure the pure vapour pressure of A is multiplied by mole fraction of A in the solution that is ${x_A}$.
Where, ${x_A} = {n_A}/({n_A} + {n_B})$
$n$ is the number of moles.
Now total vapour pressure of the solution when two liquids are added is;
${P_T} = {P_A} + {P_B} - (2)$
Now in the above equation we are provided with the total vapour pressure of the solution. When two liquids are added into it. By substituting equation 1 in two we get
${P_T} = {x_A}{P^o}_A + {x_B}{P^o}_B - (3)$
The relationship of mole fraction of a solution is
${x_A} + {x_B} = 1$
$ \Rightarrow {x_B} = 1 - {x_A} - (4)$
By substituting equation 4 in equation 3 we get;
${P_T} = {x_A}{P^o}_A + (1 - {x_A}){P^o}_B$
$ \Rightarrow $ ${P_T} = {x_A}{P^o}_A - {x_A}{P^o}_B + {P^o}_B$
$ \Rightarrow $ ${P_T} = {x_A}({P^o}_A - {P^o}_B) + {P^o}_B - (5)$
Now the question we are that
\[{\mathbf{P}} = {\mathbf{254}} - {\mathbf{119}}{{\mathbf{x}}_A}\]
By equating the above equation from equation 5 we get
${P^o}_B = 254$, and
${x_A}({P^o}_A - {P^o}_B) = - 119{x_A}$
By cancelling ${x_A}$ on both the side we get
${P^o}_A - {P^o}_B = - 119$
By putting value of ${P^0}_B$ in the equation we get
${P^0}_A - 254 = - 119$
$ \Rightarrow $ ${P^o}_A = 254 - 119$
$ \Rightarrow $ ${P^o}_A = 135$
So we get the value of pure vapour pressure of both component
Hence the correct option is C.

Note:
Always use the standard equation of Raoult’s law to solve the equation for the total vapour pressure of the solution can be found out by adding the product of mole fraction and pure vapour pressure of every liquid present in the solution.