At $80^{\circ} \mathrm{C}$, the vapour pressure of pure liquid ' $\mathrm{A}^{\prime}$ is $520 \mathrm{~mm}…

At $80^{\circ} \mathrm{C}$, the vapour pressure of pure liquid ' $\mathrm{A}^{\prime}$ is $520 \mathrm{~mm} \mathrm{Hg}$ and that of pure liquid ' $\mathrm{B}$ ' is 1000 $\mathrm{mm} \mathrm{Hg}$. If a mixture solution of ' $\mathrm{A}$ ' and ' $\mathrm{B}$ ' boils at $80^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure, the amount of ' $\mathrm{A}$ ' in the mixture is $(1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg})$
  1. $52 \mathrm{~mol}$ percent
  2. $34 \mathrm{~mol}$ percent
  3. $48 \mathrm{~mol}$ percent
  4. $50 \mathrm{~mol}$ percent

Solution

At 1 atmospheric pressure the boiling point of mixture is $80^{\circ} \mathrm{C}$.

At boiling point the vapour pressure of mixture, $\mathrm{P}_{\mathrm{T}}=1$ atmosphere $=760 \mathrm{~mm} \mathrm{Hg}$. Using the relation,
$\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{x}_{\mathrm{B}}$, we get
$P_{T}=520 x_{A}+1000\left(1-x_{A}ight)$
$\left\{\because \mathrm{P}_{\mathrm{A}}^{\mathrm{o}}=520 \mathrm{~mm} \mathrm{Hg}ight.$
$\mathrm{P}_{\mathrm{B}}^{\circ}=1000 \mathrm{~mm} \mathrm{Hg} ; \mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}=1$
or $760=520 \mathrm{x}_{\mathrm{A}}+1000-1000 \mathrm{x}_{\mathrm{A}}$ or $480 \mathrm{x}_{\mathrm{A}}=240$
or $x_{A}=\frac{240}{480}==\frac{1}{2}$ or 50 mol percent .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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