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Step-by-Step Solution
Step 1: State Raoult’s Law for Ideal Solutions
For an ideal solution containing components X and Y, the total vapor pressure $P_{\text{total}}$ at a given temperature is given by:
$P_{\text{total}} \;=\; P_X^\circ \,X_X \;+\; P_Y^\circ \,X_Y$
where $P_X^\circ$ and $P_Y^\circ$ are the vapor pressures of pure X and Y respectively, and $X_X$ and $X_Y$ are their respective mole fractions in the solution.
Step 2: Express the First Condition
According to the problem, when the solution has 1 mol of X and 3 mol of Y (total 4 moles), its vapor pressure is 550 mm Hg. Therefore:
$X_X \;=\;\dfrac{1}{4}\quad\text{and}\quad X_Y \;=\;\dfrac{3}{4}$
Hence, applying Raoult’s law:
$550\;=\;P_X^\circ \times \dfrac{1}{4} \;+\; P_Y^\circ \times \dfrac{3}{4}$
Multiplying through by 4 to simplify:
$4 \times 550\;=\;P_X^\circ \;+\; 3\,P_Y^\circ \quad...\,(i)$
Thus,
$2200\;=\;P_X^\circ \;+\; 3\,P_Y^\circ$
Step 3: Express the Second Condition
Next, 1 mol of Y is added, giving now 1 mol of X and 4 mol of Y (total 5 moles), and the vapor pressure becomes $550 + 10 = 560$ mm Hg. In this new situation:
$X_X \;=\;\dfrac{1}{5}\quad\text{and}\quad X_Y \;=\;\dfrac{4}{5}$
Hence by Raoult’s law again:
$560 \;=\; P_X^\circ \times \dfrac{1}{5} \;+\; P_Y^\circ \times \dfrac{4}{5}$
Multiplying through by 5 to simplify:
$5 \times 560\;=\;P_X^\circ \;+\; 4\,P_Y^\circ \quad...\,(ii)$
Thus,
$2800\;=\;P_X^\circ \;+\;4\,P_Y^\circ$
Step 4: Subtract the Equations to Find $P_Y^\circ$
Subtract equation (i) from equation (ii) to eliminate $P_X^\circ$:
$(2800) - (2200)\;=\;(P_X^\circ + 4\,P_Y^\circ)\;-\;(P_X^\circ + 3\,P_Y^\circ)$
Which simplifies to:
$600\;=\;P_Y^\circ$
So the vapor pressure of the pure liquid Y is $600$ mm Hg.
Step 5: Substitute $P_Y^\circ$ Back to Find $P_X^\circ$
Using $P_Y^\circ = 600$ in equation (i), for example:
$2200\;=\;P_X^\circ \;+\;3 \times 600$
$2200\;=\;P_X^\circ + 1800$
$P_X^\circ\;=\;400$
Final Answer
The vapor pressures of X and Y in their pure states are:
$P_X^\circ = 400 \text{ mm Hg and } P_Y^\circ = 600 \text{ mm Hg.}$
