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Please help! - Solutions - JEE Main-2

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)
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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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