Vapour pressure  of  liquid ‘A’ is 400mm Hg and liquid ‘B’ is 600mm Hg. If the V.P of the mixture containing 1:1 mole ratio of ‘A’ and ‘B’  is 504 mm Hg then the incorrect statement is

${\mathrm{P}}_{\mathrm{A}}^0 = 400\mathrm{mm} ; {\mathrm{P}}_{\mathrm{B}}^0 = 600\mathrm{mm}$;

$$\begin{gathered} {\mathrm{X}}_{\mathrm{A}}={\mathrm{X}}_{\mathrm{B}}=0.5; \\ {\mathrm{P}}_{\mathrm{total}}=400\left(0.5\right)+600\left(0.5\right)=500 \mathrm{mm} \mathrm{Hg}; \end{gathered}$$

But the VP of the mixture is greater(504 mm Hg) than expected from Raoults law. It shows positive deviations from Raoult’s law

$\therefore {\mathrm{\Delta H}}_{\mathrm{mix}} > 0$

${\mathrm{\Delta V}}_{\mathrm{mix}} > 0$

Boiling point  of mixture is lesser than its components;

$∆{\mathrm{G}}_{\mathrm{mix}}< 0$…..for ideal and also for non ideal solutions;