Two liquids A and B form an ideal solution. A | vedclass

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(B) For the first solution: $n_A = 3 \ mol$, $n_B = 1 \ mol$. Mole fractions are $X_A = \frac{3}{4}$ and $X_B = \frac{1}{4}$.Using Raoult's Law: $P_S

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$200$

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(B) For the first solution: $n_A = 3 \ mol$, $n_B = 1 \ mol$. Mole fractions are $X_A = \frac{3}{4}$ and $X_B = \frac{1}{4}$.Using Raoult's Law: $P_S = P_A^o X_A + P_B^o X_B$.$500 = P_A^o 	imes \frac{3}{4} + P_B^o 	imes \frac{1}{4} \implies 3 P_A^o + P_B^o = 2000$ $(I)$.After adding $1 \ mol$ of $A$: $n_A = 4 \ mol$, $n_B = 1 \ mol$. Mole fractions are $X'_A = \frac{4}{5}$ and $X'_B = \frac{1}{5}$.The new vapour pressure is $500 + 20 = 520 \ mm \ Hg$.$520 = P_A^o 	imes \frac{4}{5} + P_B^o 	imes \frac{1}{5} \implies 4 P_A^o + P_B^o = 2600$ $(II)$.Subtracting equation $(I)$ from $(II)$: $(4 P_A^o + P_B^o) - (3 P_A^o + P_B^o) = 2600 - 2000$.$P_A^o = 600 \ mm \ Hg$.Substituting $P_A^o$ in $(I)$: $3(600) + P_B^o = 2000 \implies 1800 + P_B^o = 2000 \implies P_B^o = 200 \ mm \ Hg$.

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