1 , text{mole} of each of A and B form an ide | vedclass

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(D) According to Raoult's Law,$P_{total} = P_A^0 x_A + P_B^0 x_B$.For the first case: $n_A = 1, n_B = 1$. Mole fractions are $x_A = 0.5, x_B = 0.5$. $

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$140$ and $60 \, 	ext{mm Hg}$

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(D) According to Raoult's Law,$P_{total} = P_A^0 x_A + P_B^0 x_B$.For the first case: $n_A = 1, n_B = 1$. Mole fractions are $x_A = 0.5, x_B = 0.5$. $P_{total} = 100 \, 	ext{mm Hg}$.$0.5 P_A^0 + 0.5 P_B^0 = 100 \implies P_A^0 + P_B^0 = 200 \quad (1)$.For the second case: $n_A = 1, n_B = 1 + 2 = 3$. Total moles = $4$. Mole fractions are $x_A = 0.25, x_B = 0.75$. $P_{total} = 100 - 20 = 80 \, 	ext{mm Hg}$.$0.25 P_A^0 + 0.75 P_B^0 = 80 \implies P_A^0 + 3 P_B^0 = 320 \quad (2)$.Subtracting $(1)$ from $(2)$: $(P_A^0 + 3 P_B^0) - (P_A^0 + P_B^0) = 320 - 200 \implies 2 P_B^0 = 120 \implies P_B^0 = 60 \, 	ext{mm Hg}$.Substituting $P_B^0 = 60$ into $(1)$: $P_A^0 + 60 = 200 \implies P_A^0 = 140 \, 	ext{mm Hg}$.Thus,the vapour pressures are $140 \, 	ext{mm Hg}$ and $60 \, 	ext{mm Hg}$.

$A$. 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

$1 \, \text{mole}$ of each of $A$ and $B$ form an ideal solution of vapour pressure $100 \, \text{mm Hg}$. Addition of $2 \, \text{moles}$ of $B$ to it decreases the vapour pressure by $20 \, \text{mm Hg}$. The vapour pressure of $A$ and $B$ in pure state are,respectively:

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