Pressure of · Part 3

g g mol  = . mol Moles of CHCl . g . g mol  = . mol Total number of moles = . + . = . mol CH Cl = . mol . mol = . CHCl = . – . = . Using equation ( . ), total = p + ( p – p ) x = + ( – ) × . = + . = . mm Hg (ii) Using the relation ( . ), y i = p i / p total , we can calculate the mole fraction of the components in gas phase ( y i ). CH Cl = . × mm Hg = . mm Hg CHCl = . × mm Hg = . mm Hg CH Cl y = . mm Hg/ . mm Hg = . CHCl y = . mm Hg/ . mm Hg = . Note: Since, CH Cl is a more volatile component than CHCl , [ CH Cl mm Hg and CHCl = mm Hg] and the vapour phase is also richer in CH Cl [ CH Cl y = . and CHCl y = . ], it may thus be concluded that at equilibrium, vapour phase will be always rich in the component which is more volatile.