5. Liquid C and D form a real solution at 25 °C. The vapor pressure of...

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Question

Answer 1

Answer #1

P_C⁰ = 0.75 atm

P_D⁰ = 1.5 atm

X_C = 0.5

i.e. X_D = 1-0.5 = 0.5

According to the Raoult's law: P_C = X_C.P_C⁰ = 0.25 atm

The activity (actual pressure) of C = 0.5*0.75 = 0.375 atm

The activity coefficinet of liquid C = 0.25/0.375 = 0.667

Therefore, the activity coefficient of liquid D ( $\gamma$ D) = 1-0.667 = 0.333

Now, the vapor pressure of liquid D above the solution (P_D) = X_D.P_D⁰. $\gamma$ D = 0.5*1.5*0.333 = 0.25 atm

Answer 2

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