Solutions Question 322
Question: At $ 80{}^\circ C $ , the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at $ 80{}^\circ C $ and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)
Options:
A) 52 mol percent
B) 34 mol percent
C) 48 mol percent
D) 50 mol percent
Show Answer
Answer:
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 mm Hg. Using the relation, $ P_{T}=P_A^{{}^\circ }x_{A}+P_B^{{}^\circ }x_{B} $ , we get $ P_{T},=520x_{A}+1000(1-x_{A}) $
$ \therefore P_A^{0}=520,mm,Hg, $
$ P_B^{0}=1000mmHg;x_{A}+x_{B}=1 $ or $ 760=520x_{A}+1000-1000x_{A},or,480x_{A}=240 $ or $ X_{A}=\frac{240}{480}=\frac{1}{2} $ or 50 mol percent
