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  • Try this! 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 :(Mole

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

Posted by

Sabhrant Ambastha

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