A non-ideal binary solution clearly demonstrates a positive deviation from Raoult's law. The pure eq — Solutions and Colligative Properties Chemistry Question
Question
A non-ideal binary solution clearly demonstrates a positive deviation from Raoult's law. The pure equilibrium vapour pressures are $P_A^0 = 150\text{ mm Hg}$ and $P_B^0 = 250\text{ mm Hg}$. If $2\text{ moles}$ of A and $3\text{ moles}$ of B are thoroughly mixed, the observed experimental total vapour pressure was found to be $230\text{ mm Hg}$. Calculate the exact magnitude of the deviation in total vapour pressure (in $\text{mm Hg}$) from the theoretically predicted ideal behavior.
💡 Solution & Explanation
First, calculate the theoretical ideal pressure using Raoult's law: $P_{ideal} = P_A^0 X_A + P_B^0 X_B$. Total moles = $2 + 3 = 5$. $X_A = 2/5 = 0.4$, $X_B = 3/5 = 0.6$. $P_{ideal} = (150 \times 0.4) + (250 \times 0.6) = 60 + 150 = 210\text{ mm Hg}$. The observed pressure is $230\text{ mm Hg}$. The magnitude of the positive deviation is $P_{obs} - P_{ideal} = 230 - 210 = 20\text{ mm Hg}$.