The vapor pressure of a mixture of 300 mL of | vedclass

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(A) According to Raoult's law for a mixture of two volatile liquids,the total vapor pressure $P_s$ is given by: $P_s = P_A^0 X_A + P_B^0 X_B$ Where $P

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$350$

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(A) According to Raoult's law for a mixture of two volatile liquids,the total vapor pressure $P_s$ is given by: $P_s = P_A^0 X_A + P_B^0 X_B$ Where $P_A^0$ is the vapor pressure of pure ethyl alcohol,$X_A = 0.6$ is its mole fraction,$P_B^0 = 200 \ mm$ is the vapor pressure of pure propyl alcohol,and $X_B = (1 - 0.6) = 0.4$ is its mole fraction. Substituting the values: $290 = P_A^0(0.6) + 200(0.4)$ $290 = 0.6 P_A^0 + 80$ $210 = 0.6 P_A^0$ $P_A^0 = \frac{210}{0.6} = 350 \ mm$.

The vapor pressure of a mixture of $300 \ mL$ of ethyl alcohol and propyl alcohol is $290 \ mm$. The vapor pressure of propyl alcohol is $200 \ mm$. If the mole fraction of ethyl alcohol is $0.6$,what will be its vapor pressure (in $mm$) at the same temperature?

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