question_answer

The vapour pressure lowering of 0.2 molal urea solution at 40°C is (assuming latent heat of vaporization \[(\Delta H)\] is 10 Kcal/mol.)

A) 0.2 torr

B) 0.1 torr            

C) 0.5 torr

D) 0.3 torr

Correct Answer: A

Solution :

[a] Using Clapeyron - Clausius equation first calculate vapour pressure of water of

\[\Delta H=\frac{2.303\,R{{T}_{1}}{{T}_{2}}}{{{T}_{2}}-{{T}_{1}}}\log \frac{{{P}_{2}}}{{{P}_{1}}}\]

\[{{T}_{1}}=273+{{100}^{o}}=373\,K\]            \[{{T}_{2}}=273+{{40}^{o}}=313\,K\]

\[10=\frac{2.303\times 0.002\times 313\times 373}{(313-373)}\log \frac{{{P}_{2}}}{760}\]This gives vapour pressure \[{{P}_{2}}\]  of water at \[{{40}^{o}}C=58.2\,Torr\]

By Raoult law \[\frac{\Delta P}{{{P}^{o}}}=\frac{{{w}_{1}}}{{{m}_{1}}}\times \frac{{{m}_{2}}}{{{w}_{2}}}\]

\[\frac{{{w}_{1}}}{{{m}_{1}}}=0.2\] mole urea \[{{w}_{2}}=1000\,g\,{{H}_{2}}O\]

\[m=18\,g/mol\]

\[\Delta P={{P}^{o}}\times \frac{{{w}_{1}}}{{{m}_{1}}}\times \frac{{{w}_{2}}}{{{m}_{2}}}=\frac{58.2\times 0.2\times 18}{1000}=0.20\,Torr\]