Vapour pressure of chloroform `(CHCl_(3))` and dichloromethane `(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

Molar mass of `CH_(2)Cl_(2) = 12 xx 1 xx2 + 35 . 2 xx 2 = 85 " g mol"^(-1)`
Molar mass of `CHCl_(3) = 12 xx 1 + 1 xx 1 + 35 . 5 xx 3 = 119.5 " g mol"^(-1)`
Mole of `CHCl_(3) = 12 xx 1 + 1 xx + 35 . 5 xx 3 = 119. 5 " g mol"^(-1)`
Moles of `CH_(2)Cl_(2) = (40 g )/( 85 " g mol"^(-1)) = 0.47 ` mol
Moles of `CHCl_(3) = ( 25.5 g)/(119.5 " g mol"^(-1)) = 0.213 ` mol
Total number of moles = `0.47 + 0.213 = 0.683` mol
`x_(CH_(2)Cl_(2)) = (0.47 mol)/(0.683 mol) = 0.688`
`x_(CHCl_(3)) = 1.00 - 0.688 = 0.312 `
` p_("total") = p_(1)^(0) + (p_(2)^(0) - p_(1)^(0)) x_(2) = 200 (415 - 200) xx 0.688`
` = 200 + 147 . 9 = 347 ` mm Hg
(ii) Using the relation (2.17), `y_(i) = p_(i)//p_("total")` , we can calculate the mole fraction of the components in gas phase `(y_(i))`.
`p_(CH_(2)Cl_(2)) = 0.688 xx415 " mm Hg" = 286 . 5 "mm Hg"`
` p_(CHCl_(3)) = 0.312 xx200 " mm Hg " = 62.4 ` mm Hg
` y_(CH_(2)Cl_(2)) = 285.5 " mm Hg/"347.9 " mm Hg " = 0.82`
`y_(CHCl_(3)) = 62.4 " mm Hg/"347.9 " mm Hg " = 0.18 `
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"] ` , it may thus be concluded that at equilibrium , vapour phase will be always rich in the component which is more volatile .