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Home » The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
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The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

As we know that:

P_{A}^{\circ}=450 \mathrm{~mm} \text { of } \mathrm{Hg}

P_{B}^{\circ}=700 \mathrm{~mm} of \mathrm{Hg}

P_{\text {total }}=600 \mathrm{~mm} \text { of } \mathrm{Hg}

As Raoult’s law says:

P_{A}=P_{A}^{\circ} x_{A} P_{B}=P_{B}^{\circ} x_{B}=P_{B}^{\circ}\left(1-x_{A}\right)

Hence, total pressure, P_{\text {total }}=P_{A}+P_{B}

=P_{\text {total }}=P_{A}^{\circ} x_{A}+P_{B}^{\circ}\left(1-x_{A}\right)

=P_{\text {total }}=P_{A}^{\circ} x_{A}+P_{B}^{\circ}-P_{B}^{\circ} x_{A}

\Rightarrow P_{\text {total }}=\left(P_{A}^{\circ}-P_{B}^{\circ}\right) x_{A}+P_{B}^{\circ}

=>600=(450-700) \mathrm{x}_{\mathrm{A}}+700

\Rightarrow-100=-250 \mathrm{x}_{\mathrm{A}}

\Rightarrow \mathrm{XA}=0.4

Hence, x_{B}=1-x_{A}=1-0.4=0.6

Now, P_{A}=P_{A}^{\circ} x_{A}

=450 \times 0.4=180 \mathrm{~mm} of \mathrm{Hg}

{{P}_{B}}=P_{B}^{{}^\circ }{{x}_{B}}

=700\times 0.6=420~\text{mm of Hg}

Now, in the vapour phase: Mole fraction of liquid \mathrm{A}=\frac{P_{\mathrm{A}}}{P_{A}+P_{B}}

=\frac{180}{180+420}

=\frac{180}{600}

=0.30 And, mole fraction of liquid \mathrm{B}=1-0.30=0.70

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