Chemistry, asked by laxman2001, 1 year ago

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

Answered by gadakhsanket
105

Welcome dear,

● 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..

Answered by anushkarevankar2003
41

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:

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