Question

An ideal solution was obtained mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol are $2 \cdot 619K{P_a}$ and $4 \cdot 556K{P_a}$ respectively, then the composition of vapour (in terms of mole fraction) will be:
A) $0 \cdot 635{\text{ MeOH, 0}} \cdot {\text{365 EtOH}}$
B) $0 \cdot 365{\text{ MeOH, 0}} \cdot {\text{635 EtOH}}$
C) $0 \cdot 574{\text{ MeOH, 0}} \cdot {\text{326 EtOH}}$
D) $0 \cdot 173{\text{ MeOH, 0}} \cdot {\text{827 EtOH}}$

Answer 1

Hint:The given solution is an ideal solution. One can use the relation between the partial pressure of methanol and mole fraction of methanol which is equal to the partial pressure of ethanol and mole fraction of ethanol and choose the correct values that fit the ratio.

Complete answer:
1) First of all we will discuss the relation between the two solvents which are mixed. The solvents methanol and ethanol are mixed together to form the ideal solution. As the solution is ideal, we can say that both solvents can be written as in the format below,
$\dfrac{{{\text{Partial pressure of Methanol}}}}{{{\text{Mole Fraction of Methanol}}}} = \dfrac{{{\text{Partial pressure of Ethanol}}}}{{{\text{Mole Fraction of ethanol}}}}$
We can write this equation in a simpler form as below,
$\dfrac{{{P_M}}}{{{X_M}}} = \dfrac{{{P_E}}}{{{X_E}}}$
In the above equation,
${P_M}$ is the partial pressure of methanol
${X_M}$ is the mole fraction of methanol
${P_E}$ is the partial pressure of ethanol
${X_E}$ is the mole fraction of ethanol
2) Now as we know the values of the partial pressure of methanol and the partial pressure of ethanol, we can rearrange the above equation as below,
$\dfrac{{{P_M}}}{{{P_E}}} = \dfrac{{{X_M}}}{{{X_E}}}$
As we know the values of ${P_M}$ is equal to $2 \cdot 619K{P_a}$ and ${P_E}$ is equal to $4 \cdot 556K{P_a}$, by putting these values in the above equation we get,
$\dfrac{{{P_M}}}{{{P_E}}} = \dfrac{{2 \cdot 619}}{{4 \cdot 556}}$
Now by calculating the above equation we get the value as,
$\dfrac{{{P_M}}}{{{P_E}}} = 0 \cdot 5748$
3) Now let us analyze the given options and see which option can give the ration value of $\dfrac{{{X_M}}}{{{X_E}}}$ as ${\text{0}} \cdot {\text{5748}}$.
Option A,
$\dfrac{{{X_M}}}{{{X_E}}} = \dfrac{{0 \cdot 635}}{{0 \cdot 365}} = 1 \cdot 7397$
Hence, this option is incorrect.
Option B,
$\dfrac{{{X_M}}}{{{X_E}}} = \dfrac{{0 \cdot 365}}{{0 \cdot 635}} = 0 \cdot 5748$
As the value matches with the ratio $\dfrac{{{P_M}}}{{{P_E}}}$, this option is correct.
Option C,
$\dfrac{{{X_M}}}{{{X_E}}} = \dfrac{{0 \cdot 574}}{{0 \cdot 326}} = 1 \cdot 7407$
Hence, this option is incorrect.
Option D,
$\dfrac{{{X_M}}}{{{X_E}}} = \dfrac{{0 \cdot 173}}{{0 \cdot 827}} = 0 \cdot 2091$
Hence, this option is incorrect.
As the value in option B that $0 \cdot 365{\text{ MeOH, 0}} \cdot {\text{635 EtOH}}$ matches with the $\dfrac{{{P_M}}}{{{P_E}}} = 0 \cdot 5748$, option B is the correct choice.

Note:
The given mixture of methanol and ethanol is an ideal solution and hence Raoult's law applies here. Partial pressure can be calculated as the pure vapor pressure of that component multiplied by the mole fraction of that component. The formula can be written as below,
${P_A} = P_A^0 \times {X_A}$