In a binary ideal solution of liquids A and B | vedclass

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(D) Given: $x_A = 0.7$,$x_B = 1 - 0.7 = 0.3$,$y_A = 0.4$,$y_B = 1 - 0.4 = 0.6$.According to Raoult's law,$P_A = x_A P_A^o$ and $P_B = x_B P_B^o$.The t

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$20 \, mm$ and $70 \, mm$

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(D) Given: $x_A = 0.7$,$x_B = 1 - 0.7 = 0.3$,$y_A = 0.4$,$y_B = 1 - 0.4 = 0.6$.According to Raoult's law,$P_A = x_A P_A^o$ and $P_B = x_B P_B^o$.The total pressure $P_T = P_A + P_B = x_A P_A^o + x_B P_B^o = 0.7 P_A^o + 0.3 P_B^o$.Also,for vapor phase,$y_A = \frac{P_A}{P_T} = \frac{x_A P_A^o}{P_T}$.So,$P_T = \frac{x_A P_A^o}{y_A} = \frac{0.7 P_A^o}{0.4} = 1.75 P_A^o$.Equating the two expressions for $P_T$: $1.75 P_A^o = 0.7 P_A^o + 0.3 P_B^o$.$1.05 P_A^o = 0.3 P_B^o \implies P_B^o = \frac{1.05}{0.3} P_A^o = 3.5 P_A^o$.Given $P_A^o + P_B^o = 90 \, mm$,substitute $P_B^o$: $P_A^o + 3.5 P_A^o = 90 \implies 4.5 P_A^o = 90$.$P_A^o = 20 \, mm$ and $P_B^o = 3.5 	imes 20 = 70 \, mm$.

In a binary ideal solution of liquids $A$ and $B$ at a constant temperature,the mole fraction of $A$ in the liquid phase is $0.7$ and in the vapor phase is $0.4$. If $P_A^o + P_B^o = 90 \, mm$,then the vapor pressures of pure liquids $A$ and $B$ at that temperature are respectively:

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