Correct Answer – A
`n_(CHCl_(3))=(25.5)/(119.5)=0.213`
`n_(CH_(2)Cl_(2))=(40)/(85)=0.47`
`P_(T)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)`
`=200xx(0.213)/(0.683)+41.5xx(0.47)/(0.683)`
`=62.37+28.55=90.92`

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”) = [email protected]_(3) + ([email protected] CH_(2)Cl_(2) – [email protected] 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`

(i). Molar mass of `CH_(2)Cl=12xx1+1xx2+35.5xx2=85″ g “mol^(-1)`
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+0213=0.683`mol
`x_(CH_(2)Cl_(2))=(0.47mol)/(0.683″ mol”)=0.688`
`x_(CHCl_(3))=1.00-0.688=0.312`
Using equation
`P_(“total”)=p_(1)^(0)+(p_(2)^(0)-p_(1)^(0))x_(2)=200+(415-200)xx0.688`
`=200+147.9=347.9″ mm Hg”`
(ii) Using the relation `y_(1)=P_(i)//p_(“total”)` we can calcualte the mole fraction of the components in gas phase `(y_(i))`
`Pp_(CH_(2)Cl_(2))=0.688xx415mm” 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”//347.9mm” “Hg=0.82`
`y_(CHCl_(3))=62.4″ mm Hg”//347.9″ mm Hg”=0.18`