Question

a mixture of two miscable liquids has the number of phase equal​

Answer 1

Answer:

Let XA and XB mole fraction of A and B be present in solution.

From Raoult's law

PT=PA0⋅XA+PB0⋅XB

PT=(0.4)XA+1.2(XB) ....(i)

Also, PA′=PA0⋅XA=PT×XA′

where, XA′ are mole fraction of A in gaseous phase

∴(0.4)XA=PT⋅(0.4)

∴PT=XA ....(ii)

Similarly, for XB′ mole fraction of B in gaseous phase

(1.2)XB=PT×0.6(∵XA′+XB′=1)

∴PT=2XB .....(iii)

By eqs. (i) and (ii)

Answer 2

Answer:

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Explanation:

A mixture of two miscible liquids A and B is distilled under equilibrium conditions at 1 atm pressure. The mole fraction of A in solution and vapour phase are 0.30 and 0.60 respectively. Assuming ideal behaviour of the solution and the vapour, calculate the ratio of the vapour pressure of pure A to that of pure B.