JEE Main & Advanced Chemistry Solutions / विलयन Sample Paper Topic Test - Solutions

Normal boiling point of water is 373 K (at 760 mm). Vapor pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mol, the boiling point of water at 23 mm pressure will be:

A) 250 K

B) 294 K

C) 51.6 K

D) 12.5 K

Correct Answer: B

Solution :

[b] Applying ClausiusClapeyron equation, we get

\[\log \frac{{{P}_{2}}}{{{P}_{1}}}=\frac{\Delta {{H}_{v}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}\times {{T}_{2}}} \right]\]

\[\log \frac{760}{23}=\frac{40656}{2.303\times 8.314}\left[ \frac{373-{{T}_{1}}}{373{{T}_{2}}} \right]\]

This gives \[{{T}_{1}}\] = 294.4 K.