1 mol of liquid A and 2 mol of liquid B mak | vedclass

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(C) According to Raoult's law,the ideal vapour pressure $(P_{ideal})$ is calculated as: $P_{ideal} = X_A P_A^0 + X_B P_B^0$ Given: $n_A = 1 \ mol$,$n_

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may be a minimum boiling azeotrope

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(C) According to Raoult's law,the ideal vapour pressure $(P_{ideal})$ is calculated as: $P_{ideal} = X_A P_A^0 + X_B P_B^0$ Given: $n_A = 1 \ mol$,$n_B = 2 \ mol$. Total moles = $1 + 2 = 3 \ mol$. Mole fractions: $X_A = 1/3$,$X_B = 2/3$. $P_{ideal} = (1/3 	imes 45) + (2/3 	imes 36) = 15 + 24 = 39 \ torr$. The observed vapour pressure is $42 \ torr$. Since $P_{observed} > P_{ideal}$,the solution shows positive deviation from Raoult's law. Solutions showing positive deviation exhibit an increase in volume upon mixing $(\Delta V_{mix} > 0)$ and may form a minimum boiling azeotrope.

$1 \ mol$ of liquid $A$ and $2 \ mol$ of liquid $B$ make a solution having an observed vapour pressure of $42 \ torr$. The vapour pressures of pure $A$ and pure $B$ are $45 \ torr$ and $36 \ torr$ respectively. The described solution:

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