Vapour pressure of chloroform `(CHCl_(3))` and dichloromethane `(CH_(2) Cl_(2))` at 298 K and 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 ef `CH_(2)Cl_(2)` at 298 K and (ii) mole fraction of each component in vapour phase.

Explanation

(i) Molar mass of`CH_(2)Cl_(2)`
`(12 xx 1 + 1 xx 2 + 35.5 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)`
moles of `CH_(2)Cl_(2) = (40)/(85)` = 0.47 mol
Moles of `CHCl_(3) = (25.5)/(119.5) = 0.213 ` mol
Total number of moles in the solution
= 0·47 + 0·213
= 0.683 mol
`X CH_(2) Cl_(2) = ("mole of "CH_(2) Cl_(2))/("Total mole of solutions")`
`= (0.47)/(0.683) = 0.688`
`X CH_(2) Cl_(2) = (1.00 - 0.688) = 0.312`
Now we know that
` P_("total") = P_(A)^(0) + (P_(A)^(0) - P_(A)^(0) ) xx CH_(2) Cl_(2)`
`X CH_(2) Cl_(2) = 200 + (415 - 200) xx 0.688`
= 200 + 147.9
= 347.9 mm Hg
(ii) Mole fraction of component in vapour phase
`P_(A)^(0) = CHCl_(3) `
`P_(B)^(0) = CH_(2) Cl_(2)`
`P_(CH_(2)Cl_(2)) = P_(CH_(2) xx Cl_(2))^(0) xx X_(CH_(2)Cl_(2))`
= 415 `xx `0.688
= 285.5 mm Hg
`P_(CHCl_(3)) = P_(CHCl_(3))^(0) xx X_(CHCl_(3))`
= 200 `xx 0.312`
= 62.4 mm Hg
P = y `xx P_("total")`
`Y_(CH_(2)Cl_(2)) = (P_(CH_(4)Cl_(2))/(P_("total")))`
` = (285.5)/(347.9) = 0.82`
`Y_(CHCl_(3)) = (P_(CHCl_(3))/(P_("Total")))`
`= (62.4)/(347.9) = 0.18`