A) 0.027 mmHg
B) 0.028 mmHg
C) 0.017 mmHg
D) 0.031 mmHg
Correct Answer: C
Solution :
Lowering of vapour pressure \[={{p}^{0}}-p={{p}^{0}}.{{x}_{solute}}\] \[\therefore \]\[\Delta p=35\times \frac{0.6/60}{\frac{0.6}{60}+\frac{360}{18}}\] \[=35\times \frac{.01}{.01+20}=35\times \frac{.01}{20.01}\]\[=.017mm\,Hg\]You need to login to perform this action.
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