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

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(A) According to Raoult's Law for an ideal solution,$P_s = P_X^o X_X + P_Y^o X_Y$. Initially,$n_X = 1$ and $n_Y = 3$,so $X_X = 1/4$ and $X_Y = 3/4$. $

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$400 \ mm \ Hg$ and $600 \ mm \ Hg$

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(A) According to Raoult's Law for an ideal solution,$P_s = P_X^o X_X + P_Y^o X_Y$. Initially,$n_X = 1$ and $n_Y = 3$,so $X_X = 1/4$ and $X_Y = 3/4$. $550 = P_X^o (1/4) + P_Y^o (3/4) \Rightarrow P_X^o + 3P_Y^o = 2200$ (Equation $1$). After adding $1 \ mol$ of $Y$,$n_X = 1$ and $n_Y = 4$,so $X_X = 1/5$ and $X_Y = 4/5$. The new vapor pressure is $550 + 10 = 560 \ mm \ Hg$. $560 = P_X^o (1/5) + P_Y^o (4/5) \Rightarrow P_X^o + 4P_Y^o = 2800$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(P_X^o + 4P_Y^o) - (P_X^o + 3P_Y^o) = 2800 - 2200 \Rightarrow P_Y^o = 600 \ mm \ Hg$. Substituting $P_Y^o = 600$ into Equation $1$: $P_X^o + 3(600) = 2200$ $\Rightarrow P_X^o + 1800 = 2200$ $\Rightarrow P_X^o = 400 \ mm \ Hg$.

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