A) \[52\text{ }mol\]percent
B) \[34\text{ }mol\]percent
C) \[48\text{ }mol\]percent
D) \[50\text{ }mol\]percent
Correct Answer: D
Solution :
At 1 atmospheric pressure the boiling point of mixture is\[80{}^\circ C\]. At boiling point the vapour pressure of mixture, \[{{P}_{T}}=1\]atmosphere \[=760\text{ }mmHg.\] Using the relation, \[{{P}_{T}}=p_{A}^{o}{{x}_{A}}+p_{B}^{o}{{x}_{B}},\]'we get \[{{P}_{T}}=520{{x}_{A}}+1000\,(1-{{x}_{A}})\] \[\{\because \,\,p_{A}^{o}=520\,mm\,\,Hg,\,p_{B}^{o}=1000mm\,Hg,{{x}_{A}}+{{x}_{B}}=1\}\]\[760=520{{x}_{A}}+1000-1000{{x}_{A}}\] \[{{x}_{A}}=\frac{240}{480}=\frac{1}{2}\] or \[50\,mol\] percentYou need to login to perform this action.
You will be redirected in
3 sec