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) Calculation of vapour pressure of the solution
Mass of `CHCl_(3)=25.5g`
Mass of `CH_(2)Cl_(2)=40 g`
Molar mass of `CHCl_(3)=12+1+3xx35.5="119.5 g mol"^(-1)`
Molar mass of `CH_(2)Cl_(2)=12+2+2xx35.5="85 g mol"^(-1)`
`therefore" Moles of "CHCl_(3)=("25.5 g")/("119.5 g mol"^(-1))="0.213 mole , Moles of "CH_(2)Cl_(2)=("40 g ")/("85 g mol"^(-1))="0.470 mole"`
`"Mole fraction of "CHCl_(3)(x_(CHCl_(3)))=(0.213)/(0.213+0.470)=0.312`
`"Mole fraction of "CH_(2)Cl_(2)(x_(CH_(2)Cl_(2)))=1-0.312=0.688`
`P_("total")=p_(CHCl_(3))+p_(CH_(2)Cl_(2))=x_(CHCl_(3))xxp_(CHCl_(3))^(@)+x_(CH_(2)Cl_(2))xxp_(CH_(2)Cl_(2))^(@)`
`=0.312xx200+0.688xx415=62.4+285.5=347.9" mm"`
(ii) Calculation of mole fraction of each component in vapour phase
As calculated above, `p_(CHCl_(3))="62.4 mm, "p_(CH_(2)Cl_(2))="285.5 mm, "p_("total")="347.9 mm"`
`"Mole fraction of "CHCl_(3)" in the vapour phase "(y_(CHCl_(3)))=(p_(CHCl_(3)))/(p_("total"))=("62.4 mm")/("347.9 mm")=0.18`
Mole fraction of `CH_(2)Cl_(2)` in the vapour phase `(y_(CH_(2)Cl_(2)))=1-y_(CHCl_(3))=1-0.18=0.82`
Note. It is interesting to note that `p_(CH_(2)Cl_(2))^(@) gt p_(CHCl_(3))^(@)`. The show that `CH_(2)Cl_(2)` is more volatile than `CHCl_(3)`. Also we observe that mole fraction of `CH_(2)Cl_(2)` is greater than that of `CHCl_(3)` in the vapour phase. Thus, we conclude that vapour phase is richer in the more volatile component.