Vapour pressure of chlolfom `(CHCI_(3))` and dichloromethane `(CH_(2) CI_(2))` at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate the vapour pressure of the solution prespared by miximng 25 g of `CHCI_(3)` and 45 g of `CH_(2) CI_(2)` at 298 K. Also find the mole fraction of `CHI_(3)` in the vapour phase .

Answer

Total vapour pressure=353.99 mm Hg, Mole fraction of `CHCl_(3)` in vapour phase =0.16

Explanation

Molar mass of chloroform `(CHCl_(3)) = 12 + 1 + 35 . 5 xx 3 = 119 . 5 g mol^(-1)`
Molar mass of dichloromethane `(CH_(2) Cl_(2)) = 12 + 2 + 71 = 85 g mol^(-1)` ,br> `W_("chloroform") = 25 g , W_("dichloromethane") = 45 g`
`n_("chloroform ") = ( 25 g)/( 119 . 5 g mol^(-1)) = 021`
`n_("dichoromethane ") = (45 g)/( 85 g mol^(-1)) = 0 . 53`
`x_("chloroform ") = ( 0 . 21)/( 0 . 21 + 0 .53) = (21)/( 74)`
`x_("dichloromethane ") = 1 - (21)/( 74) = (53)/( 74)`
`p_("solution ") = p_("chloroform ") + p_("dichloromethane")`
` p_("solution ") = p_("chloroform ")^(@) xx x_("chloroform")`
+ `p_("dichloromethane "^(@) xx x_("dichloromethane ")`
`= (200 xx (21)/( 74) + 415 xx (53)/(74)) ` mm Hg
`= (1)/( 74) [ 4200 + 21995]`
`= (26195)/( 74)` m m Hg
`= 353 . 99 ` m m Hg
Mole fraction of chloroform in vapour phase , `y_("chloroform") = (p_("chloroform"))/(p_("Solution "))`
`= (4200//74)/( 26195 // 74) = 0 . 16`