Vapour pressure of chloroform `(CHCl_(3))` and dichloromeane `(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.

Explanation

(i) Molar mass of `CH_(2)Cl_(2)`
`=12xx1+1xx2+35.5xx2`
`= 85 g mol^(-1)`
(ii) Molar mass of `CHCl_(3)`
`=12xx1+1xx1+35.5xx3`
`=119.5 g mol^(-1)`
Moles of `CH_(2)Cl_(2) = (40g)/(85g mol^(-1))=0.47` mol
Moles of `CHCl_(3)=(25.5g)/(119.5 g mol^(-1))=0.213` mol
Total number of moles `= 0.47+0.213`
= 0.683 mol
`chi_(CH_(2)Cl_(2))=(0.47 mol)/(0.683 mol)=0.688`
`chi_(CHCl_(3))=1.00-0.688 = 0.312`
Using equation (2.16)
`p_("total")=p_(1)^(0)+(p_(2)^(0)-p_(1)^(0))chi_(2)`
`= 200+(415-200)xx0.688`
`=200+147.9=347.9` mm Hg
(ii) Using the relation `p_(i)=y_(1)//p_("total")` we can calculate the mole :
`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_(CH_(2)Cl_(3)=62.4` mm Hg/347.9 mm Hg = 0.18
Note : since, `CH_(2)Cl_(2)` is a more 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.