Saved Bookmarks
| 1. |
Vapour pressure of chloroform (CHCl_(3)) and dichloromethane (CH_(2)Cl_(2)) at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate (i) 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) the mole fraction of each component in vapour phase. |
|
Answer» Solution :(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. |
|