At 300 K the vapour pressure of an ideal solution containing 1.0 mole each of volatile liquids X and Y is 1000 mm . Keeping the temperature constant, when 2.0 moles of liquid X is added to the solution, its vapour pressure increases by 200 mm . Calculate the vapour pressure of X and Y in their pure state

Step-by-step solution

Let P_X^0 and P_Y^0 be the vapour pressures of pure liquids X and Y respectively. According to Raoult's law, the total vapour pressure P_ total is given by P_ total = x_X P_X^0 + x_Y P_Y^0 . For the initial solution containing 1.0 mole of X and 1.0 mole of Y, the mole fractions are x_X = 0.5 and x_Y = 0.5 . Given P_ total = 1000 mm, we have: 0.5 P_X^0 + 0.5 P_Y^0 = 1000 P_X^0 + P_Y^0 = 2000 (Equation 1). After adding 2.0 moles of X, the total moles of X is 1.0 + 2.0 = 3.0 and the total moles of Y is 1.0 . The new m