# At $\dpi{100} 80^{\circ}C$ the vapour pressure of pure liquid  A is 520 mm of $\dpi{100} Hg$ and that of pure liquid B is 1000  mm of  $\dpi{100} Hg$. If a mixture solution of  A and B boils at $\dpi{100} 80^{\circ}C$ and 1 atm pressure, the amount of  A in  the mixture is $\dpi{100} \left ( 1 atm = 760 \: mm\: of Hg \right )$ Option 1) $50 \: mol \: percent$ Option 2) $52 \: mol \: percent$ Option 3) $34 \: mol \: percent$ Option 4) $48 \: mol \: percent$

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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