The vapour pressure of chloroform `(CHCl)_(3)` and dichlorocethene `(CH_(2)Cl_(2))` at `298 K` is `200 mm Hg ` and `415 mm Hg`, respectively. Calculate
a. 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`.
b. Mole fractions of each components in vapour phase .

Explanation

(i). Molar mass of `CH_(2)Cl=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)=(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+0213=0.683`mol
`x_(CH_(2)Cl_(2))=(0.47mol)/(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_(i)//p_("total")` we can calcualte the mole fraction of the components in gas phase `(y_(i))`
`Pp_(CH_(2)Cl_(2))=0.688xx415mm" 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"//347.9mm" "Hg=0.82`
`y_(CHCl_(3))=62.4" mm Hg"//347.9" mm Hg"=0.18`