1 , text{mole} of liquid A and 2 , text{moles | vedclass

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(C) Vapour pressure expected by Raoult's law is calculated as:$P_{	ext{expected}} = P_A^o X_A + P_B^o X_B = 45 	imes \frac{1}{1+2} + 36 	imes \frac

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is a minimum boiling azeotrope

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(C) Vapour pressure expected by Raoult's law is calculated as:$P_{	ext{expected}} = P_A^o X_A + P_B^o X_B = 45 	imes \frac{1}{1+2} + 36 	imes \frac{2}{1+2}$$P_{	ext{expected}} = 45 	imes \frac{1}{3} + 36 	imes \frac{2}{3} = 15 + 24 = 39 \, 	ext{torr}$.Since the observed vapour pressure $(42 \, 	ext{torr})$ is greater than the expected vapour pressure $(39 \, 	ext{torr})$,the solution exhibits positive deviation from Raoult's law.Solutions showing positive deviation form minimum boiling azeotropes at specific compositions.

Column-$I$ (Various solutions)	Column-$II$ (Their freezing point)
$a$. $0.1 \ M \ BaCl_2$ solution	$p$. $271 \ K$
$b$. $0.1 \ M \ NaCl$ solution	$q$. $270 \ K$
$c$. $0.1 \ M \ K_3[Fe(CN)_6]$ solution	$s$. $269 \ K$
$d$. $0.1 \ M \ Al_2(SO_4)_3$ solution	$r$. $268 \ K$

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

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