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