The vapour pressure of chloroform `(CHCl)_(3)` and dichlorocethene `(CH_(2)Cl_(2))` at `298 K` is `200 mm Hg ` and `415 mm Hg`, respectively. Calculate
a. The vapour pressure of the solution prepared by mixing `25.5 g` of `CHCl_(3)` and `40 g` of `CH_(2)_Cl(2)` at `298 K`.
b. Mole fractions of each components in vapour phase .

Explanation

(i) Molar mass of `CHCl_(3) = 1 xx 12 + 1 xx 1 + 3 xx 35.5`
` = 119.5 g mol^(-1)`
Molar mass of `CH_(2)Cl_(2) = 1 xx 12 + 2 xx 1 + 2 xx 35.5`
` = 85 g mol^(-1)`
Moles of `CHCl_(3) = (25.5g )/(119.5 g mol^(-1)) = 0.213` mol
Moles of `CH_(2)Cl_(2) = (40 g)/(55 g mol^(-1)) = 0.470` mol
Total number of moles ` = 0.213 + 0.470 = 0.683` mol
` x_(CHCl_(3)) = (0.213)/(0.683) = 0.312`
`X_(CH_(2)Cl_(2)) = 1.0-0.312 = 0.688`
`p_(CHCl_(3)) = p_(CHCl_(3))^(@) xx x_(CHCl_(3))`
` = 200 xx 0.312 = 62.4 g mm Hg`
`p_(CH_(2)Cl_(2)) = p_(CH_(2)Cl_(2))^(@) xx x_(CH_(2)Cl_(2))`
` = 4.15 xx 0.688 = 288.5`
`p_("total") = 62.4 + 288.5`
` = 347.9 mm Hg`
(ii) Mole fraction in vapour phase .
` y_(CHCL_(3)) = (p_(CHCl_(3)))/(p) = (62.4)/(347.9) = 0.18`
`y_(CH_(2)Cl_(2)) = (p_(CH_(2)Cl_(2))/(p)) = (288.5)/(347.9)`
` = 0.82`