At 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure of the solution increases by 25 mm of Hg, if one more mole of B is added to the above ideal solution at 300 K. Then the vapour pressure of A in its pure state is
(a) 300 mm of Hg
(b) 400 mm of Hg
(c) 500 mm of Hg
(d) 600 mm of Hg
Answers
Given:
- Number of moles of A (na) = 1
- Initial number of moles of B (n1b) = 2
- Initial vapour pressure of solution (Pi) = 500 mm of Hg
- Final number of moles of B (n2b) = 3
- Final vapour pressure of solution (Pf) = 525 mm of Hg
To find:
The vapour pressure of A in its pure state.
Solution:
- Let the vapour pressures of A and B in their pure states be Pa and Pb respectively.
- Let initial mole fractions of A and B be X1a and X1b respectively. So, X1a = na/(na+n1b) = 1/3 and X1b = 1- X1a = 2/3
- Let final mole fractions of A and B be X2a and X2b respectively. So, X2a = na/(na+n2b) = 1/4 and X2b = 1- X2a = 3/4
- Now according to Raoult's law, P = Pa*Xa + Pb*Xb
- Pa/3 + 2Pb/3 = 500 and Pa/4 + 3Pb/4 = 525
- Solving the above two equations we get, Pa = 300 mm of Hg
Answer:
The vapour pressure of A in its pure state = 300 mm of Hg.
Given :
Inital condition :
Moles of A is 1 mole .
Moles of B is 2 moles .
Pressure of system ,
Final condition :
Moles of A is 1 mole .
Moles of B is 3 moles .
Pressure of system ,
To find :
The vapour pressure of A in its pure state .
Solution :
Let , the vapour pressure of pure A and pure B is .
We know , pressure of system is given by :
Here , X is the mole fraction .
For initial condition :
...... ( 1 )
For final condition :
...... ( 2 )
Subtracting equation 1 by 2 .
We get :
Therefore ,
The vapour pressure of A in its pure state is 300 mm of Hg .
Therefore , option ( a ) is correct .
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Vapour Pressure
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