Question

If 2 moles of A and 3 moles of B are mixed to form an ideal solution and vapour pressure of A and B are 120 and 180 mm of Hg respectively. Then the composition of A and B in the vapour phase when the first traces of vapour are formed in the above case is:
(A) $X_A^1 = 0.407$
(B) $X_A^1 = 0.8$
(C) $X_A^1 = 0.109$
(D) $X_A^1 = 0.307$

Answer 1

Hint: The formula to find the mole fraction of a component is
\[{\text{Mole fraction = }}\dfrac{{{\text{Moles of component}}}}{{{\text{Total moles}}}}\]
We can say that the partial pressure of a component can be given by the following equation.
\[p = {p^ \circ }X\]

Complete step by step solution:
Here, we are given the answers in the mole fraction of A. We will first find the mole fractions of each component.
- The mole fraction can be found out by following formula
\[{\text{Mole fraction = }}\dfrac{{{\text{Moles of component}}}}{{{\text{Total moles}}}}\]
We know that total moles are 2+3=5 moles.
Now, mole fraction of A; ${X_A}$= $\dfrac{2}{5}$ and mole fraction of B; ${X_B}$=$\dfrac{3}{5}$
Now, we will find the total partial pressure due to both A and B.
- Raoult stated that the partial pressure of any component is proportional to its mole fraction. Dalton’s law of partial pressure says that the total vapor pressure of solution will be the sum of partial pressure of all the components.
So, we can write according to Dalton’s law that ${p_{total}} = {p_A} + {p_B}$
According to Raoult’s law, $p = {p^ \circ } \cdot X$ where p is the partial pressure of the component and x is its mole fraction. So, as we substitute the value of partial pressure in the Dalton’s law equation, we get
\[{p_{total}} = p_A^ \circ {X_A} + p_B^ \circ {X_B}{\text{ }}......{\text{(1)}}\]
We are given that $p_A^ \circ $ = partial pressure of pure A = 120 mm of Hg
$p_B^ \circ $ = partial pressure of pure B = 180 mm of Hg
We have already found that ${X_A}{\text{ and }}{{\text{X}}_B}$ is $\dfrac{2}{5}{\text{ and }}\dfrac{3}{5}$ respectively.
So, as we put all these values in equation (1), we get
\[{p_{total}} = (120)\left( {\dfrac{2}{5}} \right) + (180)\left( {\dfrac{3}{5}} \right)\]
\[{p_{total}} = 48 + 108 = 156mm{\text{ of Hg}}\]
Now, partial pressure of a component is related with the total pressure (${p_{total}}$) and its mole fraction as
\[p = X \cdot {p_{total}}{\text{ }}....{\text{(2)}}\]
So, for the component A, we know that the partial pressure ${p_A} = p_A^ \circ \cdot {X_A}$
So, we obtained that ${p_A} = 120 \times \dfrac{2}{5} = 48mm{\text{ of Hg}}$ and ${p_{total}}$ is 156mm of Hg. So, we can write the equation (2) for A as
\[{p_A} = {X_A} \cdot {p_{total}}\]
So,
\[48 = {X_A} \cdot 156\]
Thus, ${X_A} = \dfrac{{48}}{{156}} = 0.3076$

Therefore, the correct answer is (D).

Note: Remember that here we do not need to convert the given unit of pressure into the SI unit of pressure. We can use the value of pressure in any of the given units but we need to take care that the value of pressure of both the components is in the same unit of pressure.