Question

Assuming the formation of an ideal solution, the boiling point of a mixture containing 1560 g benzene (molar mass =78) and 1125 g chlorobenzene (molar' mass = 112.5)' using the given vapour pressure vs. temperature diagram against anexternal pressure of 1000 torr, is :

• A. ${90}^{o}C$
• B. ${100}^{o}C$
• C. ${110}^{o}C$
• D. ${120}^{o}C$

Answer 1

The correct option is B ${100}^{o}C$

${\chi }_{benzene}=\frac{1560/78}{1560/78+1125/112.5}=\frac{2}{3},{\chi }_{chlorobenzene}=\frac{1}{3}$
At boiling point, vapour pressure of solution should be 1000 torr.
At ${100}^{o}C$, from the graph,
$P_{chlorobenzene}250\times \frac{1}{3}torr$
Also, $P_{benzene}=\frac{2}{3}\times 1300$
$\therefore$ Total vapour of solution at ${100}^{o}C=\frac{250}{3}+\frac{2600}{3}=950torr$, (closest to 1000 torr)
Hence (B) is the correct answer.