The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
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`p_A^0 = 450 mm , p_B^0 = 700 mm, p_("total") = 600 mm` Applying Raoult.s law, `p_A = x_A xx p_A^0` `p_B = x_B xx p_B^0 = (1- x_A)p_B^0` `p_("total") = p_A + p_B = x_Ap_A^0 + (1-x_A)p_B^0 = p_B^0 + (p_A^0 - p_B^0)x_A` Substituting the values in the above equation, we get `600 = 700 + (450 - 700)x_A " or " 250x_A = 100 " or "x_A = (100)/(250) = 0.40` Thus, composition of the liquid mixture will be `x_A = 0.40 ` `x_B = 1-0.40 = 0.60` ` therefore p_A = x_A xx p_A^0 = 0.40 xx 450 mm = 180 mm` `p_B = x_B xx p_B^0 = 0.60 xx 700 mm = 420 mm` Mole fraction of A in the vapour phase ` = (p_A)/(p_A + p_B) = (180)/(180 + 420) = 0.30` Mole fraction of B in the vapour phase = 1 - 0.30 = 0.70.
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