Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2CI2) at 25oC are 200 mmHg and 41.5 mmHg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCI3 and 40 g of CH2CI2 at the same temperature will be :

(Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2CI2 = 85 u)

  • Option 1)

    615.0 mmHg

  • Option 2)

    347.9 mmHg

  • Option 3)

    285.5 mmHg

  • Option 4)

    173.9 mmHg

 

Answers (1)

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.

 

 Moles of CHCl_{3}=\frac{25.5}{119.5}=0.213

Moles of CH_{2}Cl_{2}=\frac{40}{85}=0.471

Mole fraction X_{CHCl_{3}}=\frac{0.213}{0.471+0.213}=\frac{0.213}{0.684}=0.311

X_{CH_{2}Cl_{2}}=\frac{0.471}{0.684}=0.689

Vapour pressure = 0.311\times P_{CHCl_{3}}+0.689\ P_{CH_{2}Cl_{2}}

                    =0.311\times 200 + 0.689 \times 41.5=90.7935\ mm Hg

The closest option is 173.9 mm Hg.

 


Option 1)

615.0 mmHg

Incorrect

Option 2)

347.9 mmHg

Incorrect

Option 3)

285.5 mmHg

Incorrect

Option 4)

173.9 mmHg

Correct

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