Solutions and Colligative Properties 1 Question 2
3. Liquid $M$ and liquid $N$ form an ideal solution. The vapour pressures of pure liquids $M$ and $N$ are 450 and $700 \mathrm{mmHg}$, respectively, at the same temperature. Then correct statement is
(2019 Main, 9 April I)
$x_{M}=$ 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) $\frac{x_{M}}{x_{N}}>\frac{y_{M}}{y_{N}}$
(b) $\frac{x_{M}}{x_{N}}=\frac{y_{M}}{y_{N}}$
(c) $\frac{x_{M}}{x_{N}}<\frac{y_{M}}{y_{N}}$
(d) $\left(x_{M}-y_{M}\right)<\left(x_{N}-y_{N}\right)$
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Solution:
- Key Idea For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. This is known as Raoult’s law.
Liquid $M$ and $N$ form an ideal solution. Vapour pressures of pure liquids $M$ and $N$ are 450 and $700 \mathrm{~mm} \mathrm{Hg}$ respectively.
$\therefore \quad p^{\mathrm{o}}{ }{N}>p^{\mathrm{o}}{ }{M}$
So, by using Raoult’s law
and $\quad x_{M}>y_{M}$
Multiplying (i) and (ii) we get
$$ \begin{array}{rlrl} & & y_{N} x_{M}>y_{M} x_{N} \ \therefore & \frac{x_{M}}{x_{N}}>\frac{y_{M}}{y_{N}} \end{array} $$
Thus, correct relation is (a).
