A solution is preparedb mixind 8.5 of g CH_(2) Cl_(2) and 11.95 g of CHCl_(3). If vapour pressure of CH_(2)Cl_(2) and CHCl_(2) and CHCl_(3) at 298 K are 415 and 200 mmHg respectively. The mole fraction of CHCl_(3) in vapour form is: (Molar mass of Cl=35.5 g mol ^(-1))

A solution is preparedb mixind 8.5 of g CH_(2) Cl_(2) and 11.95 g of C

1.	A solution is preparedb mixind 8.5 of g CH_(2) Cl_(2) and 11.95 g of CHCl_(3). If vapour pressure of CH_(2)Cl_(2) and CHCl_(2) and CHCl_(3) at 298 K are 415 and 200 mmHg respectively. The mole fraction of CHCl_(3) in vapour form is: (Molar mass of Cl=35.5 g mol ^(-1))
Answer» `0.162` `0.675` `0.325` `0.486` Solution :Molar mass os `CHCl_(3) = 119.5` g/mol Molar mass of `CH_(2) Cl_(2) =85` g/mol. molcs of `HCCl_(3 ) = (11.95)/( 119.5) =0.1` mol. Moles of `CH_(2)Cl_(2)=(8.5)/(85)=0.1` mol Molefraction of `ChCl_(3)=(0.1)/(0.2) =0.5` mol. Mole fraction of `CH_(2) Cl_(2) =(0.1)/(0.2) = 0.5` mol. Given-Vapour opressure of `CHCl_(3)=200 mm Hg = 0.263atm.` Vapour prcssure of `CH_(2) Cl_(2) =415 m m Hg = 0.546` ATM.) (1 atm = 760 mm Hg) `therefore P _(("above solution"))` = Mole fractin of `CHCl_(4)xx` (Vapour PRESSURE of `CHCl_(3)`) + Mole fraction of `CH_(2) Cl_(2)xx ` (Vapour pressure of `CH_(2) Cl_(2)`) `=0.5 xx0.263+ 0.5xx 546 =04045` Mole fraction of `CHCl_(3)` in vapour from `= (0.1315)/(0.4045) =0.325`