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

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:
$0.5 P_X^0 + 0.5 P_Y^0 = 1000 \Rightarrow P_X^0 + P_Y^0 = 2000$ (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 mole fractions are $x_X = \dfrac{3}{4} = 0.75$ and $x_Y = \dfrac{1}{4} = 0.25$. The new vapour pressure is $1000 + 200 = 1200$ mm. Thus:
$0.75 P_X^0 + 0.25 P_Y^0 = 1200 \Rightarrow 3 P_X^0 + P_Y^0 = 4800$ (Equation 2).

Subtracting Equation 1 from Equation 2:
$(3 P_X^0 + P_Y^0) - (P_X^0 + P_Y^0) = 4800 - 2000$
$2 P_X^0 = 2800 \Rightarrow P_X^0 = 1400$ mm.

Substituting $P_X^0 = 1400$ into Equation 1:
$1400 + P_Y^0 = 2000 \Rightarrow P_Y^0 = 600$ mm.

Answer: $P_X^0=1400 \quad P_Y^0=600$