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

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 g)/(85 g 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`
`chiCH_(2)Cl_(2) = 0.47 mol /0.683 mol = 0.688`
`chiCHCl_(3) = 1.00 - 0.688 = 0.312`
`P_("total") = P@CHCl_(3) + (p@ CH_(2)Cl_(2) - p@ CHCl_(3) ) chiCH_(2)Cl(2)`
`= 200 + (415 - 200 ) xx 0.688`
`= 347.9 mm Hg`
To calculate the mole fraction of component in vapour phase,
`chi_(i)^(V) = p_(i)/P_(T)`
`:. p_(CH_(2)Cl_(2)) = 0.688 xx 415 mm Hg = 285.5 mm Hg`
`p_(CHCl_(3)) = 0.312 xx 200 mm Hg = 62.4 mm Hg`
`chi_(CH_(2)Cl_(2))^V = 285.5/347.9 = 0.82`
`chi_(CHCl_(3))^V = 62.4/347.9 = 0.18`