Question

Vapour pressure of chloroform `(CHCl_(3))` and dichloromethane `(CH_(2)Cl_(2))` at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of `CHCl_(3)` and 40 g of `CH_(2)Cl_(2)` at 298 K and (ii) mole fractions of each component in vapour phase.

Answer 1

i) Molar mass of `CH_(2)Cl_(2)=12xx1+1xx2+35.5xx2=85" g mol"^(-1)`
Molar mass of `CHCl_(3)=12xx1+1xx1+35.5xx3=119.5" g mol"^(-1)`
Moles of `CH_(2)Cl_(2)=(40"g")/(85"g mol"^(-1))=0.47" mol"`
Moles of `CHCl_(3)=(25.5" g")/(119.5" g mol"^(-1))=0.213" mol"`
Total number of moles `=0.47+0.213=0.683" mol"`
`X_(CH_(2)Cl_(2))=(0.47" mol")/(0.683" mol")=0.688`
`X_(CHCl_(3))=1.00-0.688=0.312`
Using equation
`"P"_("total")=P_(1)^(0)+(P_(2)^(0)-P_(1)^(0))x_(2)=200+(415-200)xx0.688`
`=200+147.9=347.9" mm Hg"`
ii) Using the relation, `y_(1)=P_(1)//P_("total")`, we can calculate the mole fraction of the components in gas phase `(y_(1))`
`P_(CH_(2)Cl_(2))=0.688xx415" mm Hg"=285.5" mm Hg"`
`P_(CHCl_(3))=0.312xx200" mm Hg"=62.4" mm Hg"`
`y_(CH_(2)Cl_(2))=285.5" mm Hg"//347.9" mm Hg"=0.82`
`y_(CHCl_(3))=62.4" mm Hg"//347.9" mm Hg"=0.18`
Note : Since `CH_(2)Cl_(2)` is a move volatile component than `CHCl_(3)`, [`P_(CH_(2)Cl_(2))^(0)=415" mm Hg and "P_(CHCl_(3))^(0)=200" mm Hg"`] and the vapour phase is also richer in `CH_(2)Cl_(2)` [`y_(CH_(2)Cl_(2))=0.82" and "y_(CHCl_(3))=0.18`], it may thus be concluded that at equilibrium, vapour phase will be always rich in the component which is more volatile.