A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 nm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be

  • Option 1)

    360

  • Option 2)

    350

  • Option 3)

    300

  • Option 4)

    700

 

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.

 

 P=p_{A}^{0}\; \chi_{A}+p_{B}^{0}\; \chi_{B}

290=p_{A}^{0}\times 0.6+200\times (1-0.6)

290=0.6\times p_{A}^{0}+80

p_{A}^{0}=350

Correct option is 2.

 


Option 1)

360

This is an incorrect option.

Option 2)

350

This is the correct option.

Option 3)

300

This is an incorrect option.

Option 4)

700

This is an incorrect option.

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