A) 0.2
B) 0.8
C) 0.4
D) 0.5
Correct Answer: C
Solution :
| It is given that on addition of 1 mole \[{{H}_{2}}O\] in case I, V.P. increases \[{{P}_{solution}}=P_{solent}^{{}^\circ }\,\,\,{{X}_{solvent}}\] |
| Thus, we shall be having two equations relating above parameters from where \[P_{{{H}_{2}}O}^{o}\] (V.P. of pure water) can be calculated. Solute is non-volatile, hence vapour pressure of solution is |
| Let vapour pressure of \[{{H}_{2}}O=P_{{{H}_{2}}O}^{o}\] |
| And molecular weight of solute= \[{{\operatorname{m}}_{1}}\] |
| Case-1:\[{{P}_{solution}}=20.0\,Torr\], |
| \[{{n}_{1}}(solute)=\frac{6}{{{m}_{1}}}mole\] |
| \[{{n}_{2}}(solvent)=\frac{180}{18}=10mole\] |
| \[\therefore \]\[{{X}_{solvent}}=\left( \frac{10}{10+6/{{m}_{1}}} \right)=\left( \frac{10{{m}_{1}}}{10{{m}_{1}}+6} \right)\] |
| \[\therefore \]\[20=P_{{{H}_{2}}O}^{o}\left( \frac{10{{m}_{1}}}{10{{m}_{1}}+6} \right)\] |
| \[\therefore \]\[20=P_{{{H}_{2}}O}^{o}\left( \frac{10{{m}_{1}}}{10{{m}_{1}}+6} \right)\] ? (i) |
| Case-2: \[{{P}_{solution}}=20.02\,Torr,\] |
| \[{{n}_{1}}(solute)=\frac{6}{{{m}_{1}}}mole\] |
| \[{{n}_{2}}(solution)=10+1=11mole\] |
| \[{{X}_{solution}}=\left( \frac{11}{11+6/{{m}_{1}}} \right)=\left( \frac{11{{m}_{1}}}{11{{m}_{1+6}}} \right)\] |
| \[\therefore 20.02=P_{{{H}_{2}}O}^{o}\left( \frac{11{{m}_{1}}}{11{{m}_{1}}+6} \right)\] |
| Divide Eqn. (ii) by (i) |
| \[\frac{20.02}{20.0}=\frac{11}{10}\left( \frac{10{{m}_{1}}+6}{11{{m}_{1}}+6} \right)\] |
| \[\frac{20.02}{20.0}=\frac{10{{m}_{1}}+6}{11{{m}_{1}}+6}\] |
| This given \[{{\operatorname{m}}_{1}}=54gmo{{l}^{-1}}\] |
| Putting value of \[{{m}_{1}}\]into Eqs. (i) or (ii), we get |
| \[P_{{{H}_{2}}O}^{o}=20.02Torr\] |
| Ratio is \[=20.22/54=0.37\simeq 0.4\] |
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