The variation of vapour pressure $(b)$ as a function of temperature $(a)$ is studied for $\mathrm{C}_2…
The variation of vapour pressure $(b)$ as a function of temperature $(a)$ is studied for $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5, \mathrm{CCl}_4$ and $\mathrm{H}_2 \mathrm{O}$ at $760 \mathrm{~mm} \mathrm{Hg}$ and is shown in the figure below. The boiling temperatures of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5, \mathrm{CCl}_4$ and $\mathrm{H}_2 \mathrm{O}$ are 308,350 and $373 \mathrm{~K}$ respectively. Curves $A, B, C$ respectively correspond to
Given,
Boiling point (bp) at $760 \mathrm{~mm}$ of $\mathrm{Hg}$ for
(i) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5=308 \mathrm{~K}$
(ii) $\mathrm{CCl}_4=350 \mathrm{~K}$
(iii) $\mathrm{H}_2 \mathrm{O}=373 \mathrm{~K}$
$\because$ According to definition, the boiling point of a liquid is that temperature at which its vapour pressure becomes equal to the atmospheric pressure. Now, weaker is the attractive force between the molecules of given species, more is its volatile nature and smaller is its boiling point, i.e. attractive force $\propto \frac{1}{\text { volatile }}$ nature between the molecules, volatile nature $\propto$ vapour pressure over the liquid.
Temperature for boiling point, $\propto \frac{1}{\text { volatile }}$ nature.
Now, drawing perpendicular on $x$-axis, i.e. on (a) from points $B, A$ and $C$, we have
$
T_C>T_A>T_B .
$
Hence,
$
\begin{aligned}
& (A)=\mathrm{CCl}_4 \\
& (B)=\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5 \\
& (C)=\mathrm{H}_2 \mathrm{O}
\end{aligned}
$
Thus, option (c) is the correct answer
