At T(K), 2 moles of liquid A and 3 moles of l | vedclass

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(D) According to Raoult's law for an ideal solution, $P_S = X_A P_A^{\circ} + X_B P_B^{\circ}$.For the first mixture: $X_A = 2/5$, $X_B = 3/5$, and $P

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$500, 200$

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(D) According to Raoult's law for an ideal solution, $P_S = X_A P_A^{\circ} + X_B P_B^{\circ}$.For the first mixture: $X_A = 2/5$, $X_B = 3/5$, and $P_S = 320 \ mm \ Hg$.$320 = P_A^{\circ} (2/5) + P_B^{\circ} (3/5) \implies 2 P_A^{\circ} + 3 P_B^{\circ} = 1600$ ...$(I)$After adding $1$ mole of $A$ and $1$ mole of $B$, the new moles are $3$ moles of $A$ and $4$ moles of $B$. Total moles = $7$.$X_A' = 3/7$, $X_B' = 4/7$, and $P_S' = 328.6 \ mm \ Hg$.$328.6 = P_A^{\circ} (3/7) + P_B^{\circ} (4/7) \implies 3 P_A^{\circ} + 4 P_B^{\circ} = 2300.2$ ...$(II)$Solving the system of equations:Multiply $(I)$ by $3$: $6 P_A^{\circ} + 9 P_B^{\circ} = 4800$ ...$(III)$Multiply $(II)$ by $2$: $6 P_A^{\circ} + 8 P_B^{\circ} = 4600.4$ ...$(IV)$Subtracting $(IV)$ from $(III)$: $P_B^{\circ} = 199.6 \ mm \ Hg \approx 200 \ mm \ Hg$.Substituting $P_B^{\circ} = 200$ into $(I)$: $2 P_A^{\circ} + 3(200) = 1600 \implies 2 P_A^{\circ} = 1000 \implies P_A^{\circ} = 500 \ mm \ Hg$.Thus, the vapour pressures are $500 \ mm \ Hg$ and $200 \ mm \ Hg$.

$A$. Ebullioscopic constant	$I$. Depression of freezing point
$B$. Cryoscopic constant	$II$. Total pressure is the sum of partial pressures of the components
$C$. Henry's law	$III$. Elevation of boiling point
$D$. Dalton's law	$IV$. Solubility of a gas in liquid

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