A) 52 mol percent
B) 34 mol percent
C) 48 mol percent
D) 50 mol percent
Correct Answer: D
Solution :
[d] 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 mm Hg. Using the relation, \[{{P}_{T}}=P_{A}^{{}^\circ }{{x}_{A}}+P_{B}^{{}^\circ }{{x}_{B}}\], we get \[{{P}_{T}},=520{{x}_{A}}+1000(1-{{x}_{A}})\] \[\{\therefore P_{A}^{0}=520\,mm\,Hg,\] \[P_{B}^{0}=1000\text{ }mm\text{ }Hg;{{x}_{A}}+{{x}_{B}}=1\] or \[760=520{{x}_{A}}+1000-1000{{x}_{A}}\,or\,480{{x}_{A}}=240\] or\[{{X}_{A}}=\frac{240}{480}=\frac{1}{2}\] or 50 mol percentYou need to login to perform this action.
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