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  • Stuck here, help me understand: - Solutions - JEE Main-3

The vapour pressure of the solution of two liquids A(pº = 80 mm) and B(pº = 120 mm) is found to be 100 mm when X_{A} = 0.4. The result shows that

  • Option 1)

    solution exhibits ideal behaviour

  • Option 2)

    solution shows positive deviations

  • Option 3)

    solution shows negative deviations

  • Option 4)

    solution will show positive deviations for lower concentration and negative deviations for higher concentrations

 

Answers (1)

As we learned

 

Positive deviation from Raoult's Law behaviour -

When the pressure exerted by vapours of mixture is more than that in case of ideal behaviour.

 

- wherein

P_{T}> P_{A}^{0}x_{A}+ P_{B}^{0}x_{B}

 

 p_{total}=P^{0}_{A}X_{A} +P_{B}^{0}X_{B}

p_{total}=80\times 0.4+ 120\times (1-0.4)

            =32+72

P_{total}=104mm

Since P_{total}>100mm   so solution show negative deviation from Raoult’s Law.


Option 1)

solution exhibits ideal behaviour

Option 2)

solution shows positive deviations

Option 3)

solution shows negative deviations

Option 4)

solution will show positive deviations for lower concentration and negative deviations for higher concentrations

Posted by

Vakul

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