For a solution if pÅ=600 mm Hg , pB= 840 mm Hg under atmospheric conditions and vapour pressure of solution is 1 atm then find
(I) composition of solution
(ii) composition of vapour in equilibrium with solution
Answers
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● Answer -
(i) In solution, XA = 0.3 & XB = 0.7
(ii) In vapor, X'A = 0.23 & X'B = 0.77
● Explaination -
# Given -
p°A = 600 mm Hg
p°B = 840 mm Hg
P = 1 atm = 760 mm Hg
# Solution -
(i) Composition of solution -
According to Rault's law -
P = (p°A-p°B)XA + p°B
760 = (600-840)XA + 840
240XA = 80
XA = 80/240
XA = 0.3
XB = 1 - XA
XB = 1 - 0.3
XB = 0.7
(ii) Composition of vapor -
pA = XA.p°A
pA = 0.3 × 600
pA = 180 mm Hg
pB = XB.p°B
pB = 0.7 × 840
pB = 588 mm Hg
Mole fractions of components in vapor are calculated by -
X'A = pA / (pA+pB)
X'A = 180 / (180+588)
X'A = 0.23
X'B = 1 - 0.23
X'B = 0.77
Hope this helps you. Thanks for asking..
Answer: p°A= 600 mm Hg
p°B= 840 mm Hg
psol. = 1atm = 760mm Hg
(i) Composition of solution
According to Roault's law
p= (p°B- p°A). xB + p°A
760= (840- 600] xB + 600
760= 240 xB + 600
760- 600 = 240.xB
xB= 160/240
xB= 2/3
xB= 1-xA
2/3= 1- xA
xA= 1/3
(ii) Composition of vapour
pA= p°A. xA
pA= 600× 1/3
= 200 mm Hg
pB= p°B.xB
= 840× 2/3
= 280×2
= 560 mm Hg
Mole fractions are calculated by the formula,
x'A= pA/(pA+pB)
= 200/760
= 5/19
xB= 1- 5/19
= 19-5 / 19
= 14/19
Explanation: