Two liquids X and Y form an ideal solution. A | vedclass

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(B) According to Raoult's law,$P_{	ext{total}} = P_X^{\circ} X_X + P_Y^{\circ} X_Y.$For the first case: $X_X = \frac{1}{1+3} = \frac{1}{4}$ and $X_Y

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$400$ and $600$

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(B) According to Raoult's law,$P_{	ext{total}} = P_X^{\circ} X_X + P_Y^{\circ} X_Y.$For the first case: $X_X = \frac{1}{1+3} = \frac{1}{4}$ and $X_Y = \frac{3}{4}.$$550 = P_X^{\circ} 	imes \frac{1}{4} + P_Y^{\circ} 	imes \frac{3}{4} \implies P_X^{\circ} + 3P_Y^{\circ} = 2200 \dots (i)$For the second case,total moles of $Y = 3+1 = 4 \ mol.$ Total moles of solution = $1+4 = 5 \ mol.$$X_X = \frac{1}{5}$ and $X_Y = \frac{4}{5}.$New vapour pressure = $550 + 10 = 560 \ mm \ Hg.$$560 = P_X^{\circ} 	imes \frac{1}{5} + P_Y^{\circ} 	imes \frac{4}{5} \implies P_X^{\circ} + 4P_Y^{\circ} = 2800 \dots (ii)$Subtracting equation $(i)$ from $(ii)$:$(P_X^{\circ} + 4P_Y^{\circ}) - (P_X^{\circ} + 3P_Y^{\circ}) = 2800 - 2200$$P_Y^{\circ} = 600 \ mm \ Hg.$Substituting $P_Y^{\circ}$ in equation $(i)$:$P_X^{\circ} + 3(600) = 2200$$P_X^{\circ} = 2200 - 1800 = 400 \ mm \ Hg.$Thus,the vapour pressures are $400 \ mm \ Hg$ and $600 \ mm \ Hg$ respectively.

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