Answer to Question #68599 in General Chemistry for pratyush

Question

For a solution if p°a =600 mm hg, p°b =840 mm hg under atmospheric condition and vapour pressure of solution is 1 atmosphere then find question 1 composition of solution question 2 composition of vapour in equlibrium with solution

Answer 1

Use the formula of Raoult’s law
760 = (600 – 840) XA + 840
240 XA = 80
XA = 80/240 = 0.3
Use formula
XB = 1 - XA
Plug the values we get
XB= 1 − 0.4 = 0.7
Use formula
PA = PoA × XA = 600 × 0.3 = 180 mmHg
PB = PoB × XB = 840 × 0.7 = 588 mmHg
YA = 180 /(180+ 588))= 0.23
YB = 1 − YA =1 – 0 .23 = 0.77