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.

Vapour pressure of chloroform (CHCl_(3)) and dichloromeane (CH_(2)Cl_(

1.	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.
Answer» Solution :(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"]` andthe 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.