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At $80^{\circ}C$ the vapour pressure of pure liquid A is 520 mm of $Hg$ and that of pure liquid B is 1000 mm of $Hg$ . If a mixture solution of A and B boils at $80^{\circ}C$ and 1 atm pressure, the amount of A in the mixture is $\left ( 1 atm = 760 \: mm\: of Hg \right )$

Option 1)
$50 \: mol \: percent$

Option 2)
$52 \: mol \: percent$

Option 3)
$34 \: mol \: percent$

Option 4)
$48 \: mol \: percent$

Answers (1)

best_answer

As we learnt in

Rault's Law -

The total vapour pressure of binary mixture of miscible liquids be having ideally is given by

$P_{T}= P_{A}^{0}x_{A}+P_{B}^{0}x_{B}$

Where $x_{A}$ and $x_{B}$ are mole fraction of A and B in liquid phase.

- wherein

$P_{A}^{0}$ and $P_{B}^{0}$ are vapour pressures of pure liquids.

Given that $p_{A}^{0}=520\ mm Hg,\ \; p_{B}^{0}=1000\ mm Hg$

Let mole fraction of $A=\chi_{A}$ & Molefraction of $B=\chi_{B}$ at Boiling point the total pressure P is 760 mm Hg

$P=p_{A}^{0}\chi_{A}+p_{B}^{0}\chi_{B}$

$P=p_{A}^{0}\chi_{A}+p_{B}^{0}(1-\chi_{A})$

$760=520\times \chi_{A}+1000(1-\chi_{A})$

$\chi_{A}=\frac{1}{2}$ or 50 mole percent.

Correct option is 1.

Option 1)

$50 \: mol \: percent$

This is the correct option.

Option 2)

$52 \: mol \: percent$

This is an incorrect option.

Option 3)

$34 \: mol \: percent$

This is an incorrect option.

Option 4)

$48 \: mol \: percent$

This is an incorrect option.

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