Question

Liquid M and liquid N form an ideal solution. The vapour pressures of pure liquids M and N are $450$ and $700$ nm Hg, respectively at the same temperature. Then the correct statement is:

(${{X}_{M}}=$ Mole fraction of M in solution;
${{X}_{N}}=$ Mole fraction of N in solution;
${{Y}_{M}}=$ Mole fraction of M in vapour phase;
${{Y}_{N}}=$ Mole fraction of N in vapour phase)
A. $({{X}_{M}}-{{Y}_{M}})<({{X}_{N}}-{{Y}_{N}})$
B. $\dfrac{{{X}_{M}}}{{{X}_{N}}}>\dfrac{{{Y}_{M}}}{{{Y}_{N}}}$
C. $\dfrac{{{X}_{M}}}{{{X}_{N}}}<\dfrac{{{Y}_{M}}}{{{Y}_{N}}}$
D.$\dfrac{{{X}_{M}}}{{{X}_{N}}}=\dfrac{{{Y}_{M}}}{{{Y}_{N}}}$

Answer 1

For solving such types of problems, you can use Raoult’s law and Dalton’s law and compare both the laws given by them for the partial pressure of a component. You should know that, more is the vapour pressure of the component, more volatile it will be in the vapour phase.

Complete step by step solution:
Given that, Liquid M and liquid N form an ideal solution.
The vapour pressures of pure liquids M is $450$ nm Hg.
The vapour pressures of pure liquids M is $700$ nm Hg and both are in same temperatures.
${{X}_{M}}=$ Mole fraction of M in solution
${{X}_{N}}=$ Mole fraction of N in solution
${{Y}_{M}}=$ Mole fraction of M in vapour phase
${{Y}_{N}}=$ Mole fraction of N in vapour phase
Let us see what are Raoult’s law and Dalton’s law and what they say about the partial pressure. According to Raoult’s law, the partial pressure of a component in liquid phase equals the product of vapour pressure of that component in pure form and its molar fraction of the component in liquid phase. So, here the partial pressure of the component M in liquid phase (consider ${{P}_{M}}$) will be $450\times {{X}_{M}}$ and the partial pressure of the component N (consider ${{P}_{N}}$) will be $700\times {{X}_{N}}$. So, $\dfrac{{{P}_{M}}}{{{P}_{N}}}=\dfrac{450\times {{X}_{M}}}{700\times {{X}_{N}}}$. While, according to Dalton’s law, when a component is in vapour phase, its partial pressure equals the product of the total pressure (say ${{P}_{T}}$) and mole fraction of the component in vapour phase. Thus, the partial pressure of the component M in vapour phase (${{P}_{M}}$) will be ${{P}_{T}}\times {{Y}_{M}}$ and that of component N (${{P}_{N}}$) will be ${{P}_{T}}\times {{Y}_{N}}$. So, $\dfrac{{{P}_{M}}}{{{P}_{N}}}=\dfrac{{{P}_{T}}\times {{Y}_{M}}}{{{P}_{T}}\times {{Y}_{N}}}$
So, now comparing these two laws we get:
$\dfrac{{{P}_{M}}}{{{P}_{N}}}=\dfrac{450\times {{X}_{M}}}{700\times {{X}_{N}}}=\dfrac{{{P}_{T}}\times {{Y}_{M}}}{{{P}_{T}}\times {{Y}_{N}}}$
Then, $\dfrac{450{{X}_{M}}}{700{{X}_{N}}}=\dfrac{{{Y}_{M}}}{{{Y}_{N}}}$
As, the value on the left-hand side will be less than one, so the value for $\dfrac{{{Y}_{M}}}{{{Y}_{N}}}$ will be less than that of $\dfrac{{{X}_{M}}}{{{X}_{N}}}$.
So, $\dfrac{{{X}_{M}}}{{{X}_{N}}}>\dfrac{{{Y}_{M}}}{{{Y}_{N}}}$.

Hence, the correct option is B.

Note: It is important to note down that, a component having greater value for the vapour pressure in its pure form will have higher volatility in vapour phase and lesser volatility in the liquid phase. Here, in the given question the component N will be more volatile in vapour phase.